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Factorisation
Question 1 of 40
Evaluate \begin{equation*} \frac{(2^{1} + 1)(2^{2} + 1)(2^{4} + 1)(2^{8} + 1)}{(2^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{16}-1)=(2^{8}-1)(2^{8}+1) \end{equation} Cancel $(2^{8}+1)$ both in the numerator and the denominator. \begin{equation} (2^{8}-1)=(2^{4}-1)(2^{4}+1) \end{equation} Cancel $(2^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{16}-1)=(2^{8}-1)(2^{8}+1) \end{equation} Cancel $(2^{8}+1)$ both in the numerator and the denominator. \begin{equation} (2^{8}-1)=(2^{4}-1)(2^{4}+1) \end{equation} Cancel $(2^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Evaluate \begin{equation*} \frac{(2^{1} + 1)(2^{2} + 1)(2^{4} + 1)(2^{8} + 1)(2^{16} + 1)}{(2^{32}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{32}-1)=(2^{16}-1)(2^{16}+1) \end{equation} Cancel $(2^{16}+1)$ both in the numerator and the denominator. \begin{equation} (2^{16}-1)=(2^{8}-1)(2^{8}+1) \end{equation} Cancel $(2^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Evaluate \begin{equation*} \frac{(2^{1} + 1)(2^{2} + 1)(2^{4} + 1)(2^{8} + 1)(2^{16} + 1)(2^{32} + 1)}{(2^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{64}-1)=(2^{32}-1)(2^{32}+1) \end{equation} Cancel $(2^{32}+1)$ both in the numerator and the denominator. \begin{equation} (2^{32}-1)=(2^{16}-1)(2^{16}+1) \end{equation} Cancel $(2^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{64}-1)=(2^{32}-1)(2^{32}+1) \end{equation} Cancel $(2^{32}+1)$ both in the numerator and the denominator. \begin{equation} (2^{32}-1)=(2^{16}-1)(2^{16}+1) \end{equation} Cancel $(2^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Evaluate \begin{equation*} \frac{(2^{1} + 1)(2^{2} + 1)(2^{4} + 1)(2^{8} + 1)(2^{16} + 1)(2^{32} + 1)(2^{64} + 1)}{(2^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{128}-1)=(2^{64}-1)(2^{64}+1) \end{equation} Cancel $(2^{64}+1)$ both in the numerator and the denominator. \begin{equation} (2^{64}-1)=(2^{32}-1)(2^{32}+1) \end{equation} Cancel $(2^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (2^{128}-1)=(2^{64}-1)(2^{64}+1) \end{equation} Cancel $(2^{64}+1)$ both in the numerator and the denominator. \begin{equation} (2^{64}-1)=(2^{32}-1)(2^{32}+1) \end{equation} Cancel $(2^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({2}-1)}=1$.
Evaluate \begin{equation*} \frac{(3^{1} + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)}{(3^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{16}-1)=(3^{8}-1)(3^{8}+1) \end{equation} Cancel $(3^{8}+1)$ both in the numerator and the denominator. \begin{equation} (3^{8}-1)=(3^{4}-1)(3^{4}+1) \end{equation} Cancel $(3^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{16}-1)=(3^{8}-1)(3^{8}+1) \end{equation} Cancel $(3^{8}+1)$ both in the numerator and the denominator. \begin{equation} (3^{8}-1)=(3^{4}-1)(3^{4}+1) \end{equation} Cancel $(3^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Evaluate \begin{equation*} \frac{(3^{1} + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)}{(3^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{32}-1)=(3^{16}-1)(3^{16}+1) \end{equation} Cancel $(3^{16}+1)$ both in the numerator and the denominator. \begin{equation} (3^{16}-1)=(3^{8}-1)(3^{8}+1) \end{equation} Cancel $(3^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{32}-1)=(3^{16}-1)(3^{16}+1) \end{equation} Cancel $(3^{16}+1)$ both in the numerator and the denominator. \begin{equation} (3^{16}-1)=(3^{8}-1)(3^{8}+1) \end{equation} Cancel $(3^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Evaluate \begin{equation*} \frac{(3^{1} + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)(3^{32} + 1)}{(3^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{64}-1)=(3^{32}-1)(3^{32}+1) \end{equation} Cancel $(3^{32}+1)$ both in the numerator and the denominator. \begin{equation} (3^{32}-1)=(3^{16}-1)(3^{16}+1) \end{equation} Cancel $(3^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{64}-1)=(3^{32}-1)(3^{32}+1) \end{equation} Cancel $(3^{32}+1)$ both in the numerator and the denominator. \begin{equation} (3^{32}-1)=(3^{16}-1)(3^{16}+1) \end{equation} Cancel $(3^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Evaluate \begin{equation*} \frac{(3^{1} + 1)(3^{2} + 1)(3^{4} + 1)(3^{8} + 1)(3^{16} + 1)(3^{32} + 1)(3^{64} + 1)}{(3^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{128}-1)=(3^{64}-1)(3^{64}+1) \end{equation} Cancel $(3^{64}+1)$ both in the numerator and the denominator. \begin{equation} (3^{64}-1)=(3^{32}-1)(3^{32}+1) \end{equation} Cancel $(3^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (3^{128}-1)=(3^{64}-1)(3^{64}+1) \end{equation} Cancel $(3^{64}+1)$ both in the numerator and the denominator. \begin{equation} (3^{64}-1)=(3^{32}-1)(3^{32}+1) \end{equation} Cancel $(3^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({3}-1)}=\frac{1}{2}$.
Evaluate \begin{equation*} \frac{(4^{1} + 1)(4^{2} + 1)(4^{4} + 1)(4^{8} + 1)}{(4^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{16}-1)=(4^{8}-1)(4^{8}+1) \end{equation} Cancel $(4^{8}+1)$ both in the numerator and the denominator. \begin{equation} (4^{8}-1)=(4^{4}-1)(4^{4}+1) \end{equation} Cancel $(4^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{16}-1)=(4^{8}-1)(4^{8}+1) \end{equation} Cancel $(4^{8}+1)$ both in the numerator and the denominator. \begin{equation} (4^{8}-1)=(4^{4}-1)(4^{4}+1) \end{equation} Cancel $(4^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Evaluate \begin{equation*} \frac{(4^{1} + 1)(4^{2} + 1)(4^{4} + 1)(4^{8} + 1)(4^{16} + 1)}{(4^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{32}-1)=(4^{16}-1)(4^{16}+1) \end{equation} Cancel $(4^{16}+1)$ both in the numerator and the denominator. \begin{equation} (4^{16}-1)=(4^{8}-1)(4^{8}+1) \end{equation} Cancel $(4^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{32}-1)=(4^{16}-1)(4^{16}+1) \end{equation} Cancel $(4^{16}+1)$ both in the numerator and the denominator. \begin{equation} (4^{16}-1)=(4^{8}-1)(4^{8}+1) \end{equation} Cancel $(4^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Evaluate \begin{equation*} \frac{(4^{1} + 1)(4^{2} + 1)(4^{4} + 1)(4^{8} + 1)(4^{16} + 1)(4^{32} + 1)}{(4^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{64}-1)=(4^{32}-1)(4^{32}+1) \end{equation} Cancel $(4^{32}+1)$ both in the numerator and the denominator. \begin{equation} (4^{32}-1)=(4^{16}-1)(4^{16}+1) \end{equation} Cancel $(4^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{64}-1)=(4^{32}-1)(4^{32}+1) \end{equation} Cancel $(4^{32}+1)$ both in the numerator and the denominator. \begin{equation} (4^{32}-1)=(4^{16}-1)(4^{16}+1) \end{equation} Cancel $(4^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Evaluate \begin{equation*} \frac{(4^{1} + 1)(4^{2} + 1)(4^{4} + 1)(4^{8} + 1)(4^{16} + 1)(4^{32} + 1)(4^{64} + 1)}{(4^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{128}-1)=(4^{64}-1)(4^{64}+1) \end{equation} Cancel $(4^{64}+1)$ both in the numerator and the denominator. \begin{equation} (4^{64}-1)=(4^{32}-1)(4^{32}+1) \end{equation} Cancel $(4^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (4^{128}-1)=(4^{64}-1)(4^{64}+1) \end{equation} Cancel $(4^{64}+1)$ both in the numerator and the denominator. \begin{equation} (4^{64}-1)=(4^{32}-1)(4^{32}+1) \end{equation} Cancel $(4^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({4}-1)}=\frac{1}{3}$.
Evaluate \begin{equation*} \frac{(5^{1} + 1)(5^{2} + 1)(5^{4} + 1)(5^{8} + 1)}{(5^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{16}-1)=(5^{8}-1)(5^{8}+1) \end{equation} Cancel $(5^{8}+1)$ both in the numerator and the denominator. \begin{equation} (5^{8}-1)=(5^{4}-1)(5^{4}+1) \end{equation} Cancel $(5^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{16}-1)=(5^{8}-1)(5^{8}+1) \end{equation} Cancel $(5^{8}+1)$ both in the numerator and the denominator. \begin{equation} (5^{8}-1)=(5^{4}-1)(5^{4}+1) \end{equation} Cancel $(5^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Evaluate \begin{equation*} \frac{(5^{1} + 1)(5^{2} + 1)(5^{4} + 1)(5^{8} + 1)(5^{16} + 1)}{(5^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{32}-1)=(5^{16}-1)(5^{16}+1) \end{equation} Cancel $(5^{16}+1)$ both in the numerator and the denominator. \begin{equation} (5^{16}-1)=(5^{8}-1)(5^{8}+1) \end{equation} Cancel $(5^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{32}-1)=(5^{16}-1)(5^{16}+1) \end{equation} Cancel $(5^{16}+1)$ both in the numerator and the denominator. \begin{equation} (5^{16}-1)=(5^{8}-1)(5^{8}+1) \end{equation} Cancel $(5^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Evaluate \begin{equation*} \frac{(5^{1} + 1)(5^{2} + 1)(5^{4} + 1)(5^{8} + 1)(5^{16} + 1)(5^{32} + 1)}{(5^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{64}-1)=(5^{32}-1)(5^{32}+1) \end{equation} Cancel $(5^{32}+1)$ both in the numerator and the denominator. \begin{equation} (5^{32}-1)=(5^{16}-1)(5^{16}+1) \end{equation} Cancel $(5^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{64}-1)=(5^{32}-1)(5^{32}+1) \end{equation} Cancel $(5^{32}+1)$ both in the numerator and the denominator. \begin{equation} (5^{32}-1)=(5^{16}-1)(5^{16}+1) \end{equation} Cancel $(5^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Evaluate \begin{equation*} \frac{(5^{1} + 1)(5^{2} + 1)(5^{4} + 1)(5^{8} + 1)(5^{16} + 1)(5^{32} + 1)(5^{64} + 1)}{(5^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{128}-1)=(5^{64}-1)(5^{64}+1) \end{equation} Cancel $(5^{64}+1)$ both in the numerator and the denominator. \begin{equation} (5^{64}-1)=(5^{32}-1)(5^{32}+1) \end{equation} Cancel $(5^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (5^{128}-1)=(5^{64}-1)(5^{64}+1) \end{equation} Cancel $(5^{64}+1)$ both in the numerator and the denominator. \begin{equation} (5^{64}-1)=(5^{32}-1)(5^{32}+1) \end{equation} Cancel $(5^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({5}-1)}=\frac{1}{4}$.
Evaluate \begin{equation*} \frac{(6^{1} + 1)(6^{2} + 1)(6^{4} + 1)(6^{8} + 1)}{(6^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{16}-1)=(6^{8}-1)(6^{8}+1) \end{equation} Cancel $(6^{8}+1)$ both in the numerator and the denominator. \begin{equation} (6^{8}-1)=(6^{4}-1)(6^{4}+1) \end{equation} Cancel $(6^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{16}-1)=(6^{8}-1)(6^{8}+1) \end{equation} Cancel $(6^{8}+1)$ both in the numerator and the denominator. \begin{equation} (6^{8}-1)=(6^{4}-1)(6^{4}+1) \end{equation} Cancel $(6^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Evaluate \begin{equation*} \frac{(6^{1} + 1)(6^{2} + 1)(6^{4} + 1)(6^{8} + 1)(6^{16} + 1)}{(6^{32}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{32}-1)=(6^{16}-1)(6^{16}+1) \end{equation} Cancel $(6^{16}+1)$ both in the numerator and the denominator. \begin{equation} (6^{16}-1)=(6^{8}-1)(6^{8}+1) \end{equation} Cancel $(6^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{32}-1)=(6^{16}-1)(6^{16}+1) \end{equation} Cancel $(6^{16}+1)$ both in the numerator and the denominator. \begin{equation} (6^{16}-1)=(6^{8}-1)(6^{8}+1) \end{equation} Cancel $(6^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Evaluate \begin{equation*} \frac{(6^{1} + 1)(6^{2} + 1)(6^{4} + 1)(6^{8} + 1)(6^{16} + 1)(6^{32} + 1)}{(6^{64}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{64}-1)=(6^{32}-1)(6^{32}+1) \end{equation} Cancel $(6^{32}+1)$ both in the numerator and the denominator. \begin{equation} (6^{32}-1)=(6^{16}-1)(6^{16}+1) \end{equation} Cancel $(6^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{64}-1)=(6^{32}-1)(6^{32}+1) \end{equation} Cancel $(6^{32}+1)$ both in the numerator and the denominator. \begin{equation} (6^{32}-1)=(6^{16}-1)(6^{16}+1) \end{equation} Cancel $(6^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Evaluate \begin{equation*} \frac{(6^{1} + 1)(6^{2} + 1)(6^{4} + 1)(6^{8} + 1)(6^{16} + 1)(6^{32} + 1)(6^{64} + 1)}{(6^{128}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{128}-1)=(6^{64}-1)(6^{64}+1) \end{equation} Cancel $(6^{64}+1)$ both in the numerator and the denominator. \begin{equation} (6^{64}-1)=(6^{32}-1)(6^{32}+1) \end{equation} Cancel $(6^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (6^{128}-1)=(6^{64}-1)(6^{64}+1) \end{equation} Cancel $(6^{64}+1)$ both in the numerator and the denominator. \begin{equation} (6^{64}-1)=(6^{32}-1)(6^{32}+1) \end{equation} Cancel $(6^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({6}-1)}=\frac{1}{5}$.
Evaluate \begin{equation*} \frac{(7^{1} + 1)(7^{2} + 1)(7^{4} + 1)(7^{8} + 1)}{(7^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{16}-1)=(7^{8}-1)(7^{8}+1) \end{equation} Cancel $(7^{8}+1)$ both in the numerator and the denominator. \begin{equation} (7^{8}-1)=(7^{4}-1)(7^{4}+1) \end{equation} Cancel $(7^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{16}-1)=(7^{8}-1)(7^{8}+1) \end{equation} Cancel $(7^{8}+1)$ both in the numerator and the denominator. \begin{equation} (7^{8}-1)=(7^{4}-1)(7^{4}+1) \end{equation} Cancel $(7^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Evaluate \begin{equation*} \frac{(7^{1} + 1)(7^{2} + 1)(7^{4} + 1)(7^{8} + 1)(7^{16} + 1)}{(7^{32}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{32}-1)=(7^{16}-1)(7^{16}+1) \end{equation} Cancel $(7^{16}+1)$ both in the numerator and the denominator. \begin{equation} (7^{16}-1)=(7^{8}-1)(7^{8}+1) \end{equation} Cancel $(7^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{32}-1)=(7^{16}-1)(7^{16}+1) \end{equation} Cancel $(7^{16}+1)$ both in the numerator and the denominator. \begin{equation} (7^{16}-1)=(7^{8}-1)(7^{8}+1) \end{equation} Cancel $(7^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Evaluate \begin{equation*} \frac{(7^{1} + 1)(7^{2} + 1)(7^{4} + 1)(7^{8} + 1)(7^{16} + 1)(7^{32} + 1)}{(7^{64}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{64}-1)=(7^{32}-1)(7^{32}+1) \end{equation} Cancel $(7^{32}+1)$ both in the numerator and the denominator. \begin{equation} (7^{32}-1)=(7^{16}-1)(7^{16}+1) \end{equation} Cancel $(7^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{64}-1)=(7^{32}-1)(7^{32}+1) \end{equation} Cancel $(7^{32}+1)$ both in the numerator and the denominator. \begin{equation} (7^{32}-1)=(7^{16}-1)(7^{16}+1) \end{equation} Cancel $(7^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Evaluate \begin{equation*} \frac{(7^{1} + 1)(7^{2} + 1)(7^{4} + 1)(7^{8} + 1)(7^{16} + 1)(7^{32} + 1)(7^{64} + 1)}{(7^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{128}-1)=(7^{64}-1)(7^{64}+1) \end{equation} Cancel $(7^{64}+1)$ both in the numerator and the denominator. \begin{equation} (7^{64}-1)=(7^{32}-1)(7^{32}+1) \end{equation} Cancel $(7^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (7^{128}-1)=(7^{64}-1)(7^{64}+1) \end{equation} Cancel $(7^{64}+1)$ both in the numerator and the denominator. \begin{equation} (7^{64}-1)=(7^{32}-1)(7^{32}+1) \end{equation} Cancel $(7^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({7}-1)}=\frac{1}{6}$.
Evaluate \begin{equation*} \frac{(8^{1} + 1)(8^{2} + 1)(8^{4} + 1)(8^{8} + 1)}{(8^{16}-1)} \end{equation*}
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Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{16}-1)=(8^{8}-1)(8^{8}+1) \end{equation} Cancel $(8^{8}+1)$ both in the numerator and the denominator. \begin{equation} (8^{8}-1)=(8^{4}-1)(8^{4}+1) \end{equation} Cancel $(8^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{16}-1)=(8^{8}-1)(8^{8}+1) \end{equation} Cancel $(8^{8}+1)$ both in the numerator and the denominator. \begin{equation} (8^{8}-1)=(8^{4}-1)(8^{4}+1) \end{equation} Cancel $(8^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Evaluate \begin{equation*} \frac{(8^{1} + 1)(8^{2} + 1)(8^{4} + 1)(8^{8} + 1)(8^{16} + 1)}{(8^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{32}-1)=(8^{16}-1)(8^{16}+1) \end{equation} Cancel $(8^{16}+1)$ both in the numerator and the denominator. \begin{equation} (8^{16}-1)=(8^{8}-1)(8^{8}+1) \end{equation} Cancel $(8^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{32}-1)=(8^{16}-1)(8^{16}+1) \end{equation} Cancel $(8^{16}+1)$ both in the numerator and the denominator. \begin{equation} (8^{16}-1)=(8^{8}-1)(8^{8}+1) \end{equation} Cancel $(8^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Evaluate \begin{equation*} \frac{(8^{1} + 1)(8^{2} + 1)(8^{4} + 1)(8^{8} + 1)(8^{16} + 1)(8^{32} + 1)}{(8^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{64}-1)=(8^{32}-1)(8^{32}+1) \end{equation} Cancel $(8^{32}+1)$ both in the numerator and the denominator. \begin{equation} (8^{32}-1)=(8^{16}-1)(8^{16}+1) \end{equation} Cancel $(8^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{64}-1)=(8^{32}-1)(8^{32}+1) \end{equation} Cancel $(8^{32}+1)$ both in the numerator and the denominator. \begin{equation} (8^{32}-1)=(8^{16}-1)(8^{16}+1) \end{equation} Cancel $(8^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Evaluate \begin{equation*} \frac{(8^{1} + 1)(8^{2} + 1)(8^{4} + 1)(8^{8} + 1)(8^{16} + 1)(8^{32} + 1)(8^{64} + 1)}{(8^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{128}-1)=(8^{64}-1)(8^{64}+1) \end{equation} Cancel $(8^{64}+1)$ both in the numerator and the denominator. \begin{equation} (8^{64}-1)=(8^{32}-1)(8^{32}+1) \end{equation} Cancel $(8^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (8^{128}-1)=(8^{64}-1)(8^{64}+1) \end{equation} Cancel $(8^{64}+1)$ both in the numerator and the denominator. \begin{equation} (8^{64}-1)=(8^{32}-1)(8^{32}+1) \end{equation} Cancel $(8^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({8}-1)}=\frac{1}{7}$.
Evaluate \begin{equation*} \frac{(9^{1} + 1)(9^{2} + 1)(9^{4} + 1)(9^{8} + 1)}{(9^{16}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{16}-1)=(9^{8}-1)(9^{8}+1) \end{equation} Cancel $(9^{8}+1)$ both in the numerator and the denominator. \begin{equation} (9^{8}-1)=(9^{4}-1)(9^{4}+1) \end{equation} Cancel $(9^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{16}-1)=(9^{8}-1)(9^{8}+1) \end{equation} Cancel $(9^{8}+1)$ both in the numerator and the denominator. \begin{equation} (9^{8}-1)=(9^{4}-1)(9^{4}+1) \end{equation} Cancel $(9^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Evaluate \begin{equation*} \frac{(9^{1} + 1)(9^{2} + 1)(9^{4} + 1)(9^{8} + 1)(9^{16} + 1)}{(9^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{32}-1)=(9^{16}-1)(9^{16}+1) \end{equation} Cancel $(9^{16}+1)$ both in the numerator and the denominator. \begin{equation} (9^{16}-1)=(9^{8}-1)(9^{8}+1) \end{equation} Cancel $(9^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{32}-1)=(9^{16}-1)(9^{16}+1) \end{equation} Cancel $(9^{16}+1)$ both in the numerator and the denominator. \begin{equation} (9^{16}-1)=(9^{8}-1)(9^{8}+1) \end{equation} Cancel $(9^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Evaluate \begin{equation*} \frac{(9^{1} + 1)(9^{2} + 1)(9^{4} + 1)(9^{8} + 1)(9^{16} + 1)(9^{32} + 1)}{(9^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{64}-1)=(9^{32}-1)(9^{32}+1) \end{equation} Cancel $(9^{32}+1)$ both in the numerator and the denominator. \begin{equation} (9^{32}-1)=(9^{16}-1)(9^{16}+1) \end{equation} Cancel $(9^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{64}-1)=(9^{32}-1)(9^{32}+1) \end{equation} Cancel $(9^{32}+1)$ both in the numerator and the denominator. \begin{equation} (9^{32}-1)=(9^{16}-1)(9^{16}+1) \end{equation} Cancel $(9^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Evaluate \begin{equation*} \frac{(9^{1} + 1)(9^{2} + 1)(9^{4} + 1)(9^{8} + 1)(9^{16} + 1)(9^{32} + 1)(9^{64} + 1)}{(9^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{128}-1)=(9^{64}-1)(9^{64}+1) \end{equation} Cancel $(9^{64}+1)$ both in the numerator and the denominator. \begin{equation} (9^{64}-1)=(9^{32}-1)(9^{32}+1) \end{equation} Cancel $(9^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (9^{128}-1)=(9^{64}-1)(9^{64}+1) \end{equation} Cancel $(9^{64}+1)$ both in the numerator and the denominator. \begin{equation} (9^{64}-1)=(9^{32}-1)(9^{32}+1) \end{equation} Cancel $(9^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({9}-1)}=\frac{1}{8}$.
Evaluate \begin{equation*} \frac{(10^{1} + 1)(10^{2} + 1)(10^{4} + 1)(10^{8} + 1)}{(10^{16}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{16}-1)=(10^{8}-1)(10^{8}+1) \end{equation} Cancel $(10^{8}+1)$ both in the numerator and the denominator. \begin{equation} (10^{8}-1)=(10^{4}-1)(10^{4}+1) \end{equation} Cancel $(10^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{16}-1)=(10^{8}-1)(10^{8}+1) \end{equation} Cancel $(10^{8}+1)$ both in the numerator and the denominator. \begin{equation} (10^{8}-1)=(10^{4}-1)(10^{4}+1) \end{equation} Cancel $(10^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Evaluate \begin{equation*} \frac{(10^{1} + 1)(10^{2} + 1)(10^{4} + 1)(10^{8} + 1)(10^{16} + 1)}{(10^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{32}-1)=(10^{16}-1)(10^{16}+1) \end{equation} Cancel $(10^{16}+1)$ both in the numerator and the denominator. \begin{equation} (10^{16}-1)=(10^{8}-1)(10^{8}+1) \end{equation} Cancel $(10^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{32}-1)=(10^{16}-1)(10^{16}+1) \end{equation} Cancel $(10^{16}+1)$ both in the numerator and the denominator. \begin{equation} (10^{16}-1)=(10^{8}-1)(10^{8}+1) \end{equation} Cancel $(10^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Evaluate \begin{equation*} \frac{(10^{1} + 1)(10^{2} + 1)(10^{4} + 1)(10^{8} + 1)(10^{16} + 1)(10^{32} + 1)}{(10^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{64}-1)=(10^{32}-1)(10^{32}+1) \end{equation} Cancel $(10^{32}+1)$ both in the numerator and the denominator. \begin{equation} (10^{32}-1)=(10^{16}-1)(10^{16}+1) \end{equation} Cancel $(10^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{64}-1)=(10^{32}-1)(10^{32}+1) \end{equation} Cancel $(10^{32}+1)$ both in the numerator and the denominator. \begin{equation} (10^{32}-1)=(10^{16}-1)(10^{16}+1) \end{equation} Cancel $(10^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Evaluate \begin{equation*} \frac{(10^{1} + 1)(10^{2} + 1)(10^{4} + 1)(10^{8} + 1)(10^{16} + 1)(10^{32} + 1)(10^{64} + 1)}{(10^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{128}-1)=(10^{64}-1)(10^{64}+1) \end{equation} Cancel $(10^{64}+1)$ both in the numerator and the denominator. \begin{equation} (10^{64}-1)=(10^{32}-1)(10^{32}+1) \end{equation} Cancel $(10^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (10^{128}-1)=(10^{64}-1)(10^{64}+1) \end{equation} Cancel $(10^{64}+1)$ both in the numerator and the denominator. \begin{equation} (10^{64}-1)=(10^{32}-1)(10^{32}+1) \end{equation} Cancel $(10^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({10}-1)}=\frac{1}{9}$.
Evaluate \begin{equation*} \frac{(11^{1} + 1)(11^{2} + 1)(11^{4} + 1)(11^{8} + 1)}{(11^{16}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{16}-1)=(11^{8}-1)(11^{8}+1) \end{equation} Cancel $(11^{8}+1)$ both in the numerator and the denominator. \begin{equation} (11^{8}-1)=(11^{4}-1)(11^{4}+1) \end{equation} Cancel $(11^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{16}-1)=(11^{8}-1)(11^{8}+1) \end{equation} Cancel $(11^{8}+1)$ both in the numerator and the denominator. \begin{equation} (11^{8}-1)=(11^{4}-1)(11^{4}+1) \end{equation} Cancel $(11^{4}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Evaluate \begin{equation*} \frac{(11^{1} + 1)(11^{2} + 1)(11^{4} + 1)(11^{8} + 1)(11^{16} + 1)}{(11^{32}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{32}-1)=(11^{16}-1)(11^{16}+1) \end{equation} Cancel $(11^{16}+1)$ both in the numerator and the denominator. \begin{equation} (11^{16}-1)=(11^{8}-1)(11^{8}+1) \end{equation} Cancel $(11^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{32}-1)=(11^{16}-1)(11^{16}+1) \end{equation} Cancel $(11^{16}+1)$ both in the numerator and the denominator. \begin{equation} (11^{16}-1)=(11^{8}-1)(11^{8}+1) \end{equation} Cancel $(11^{8}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Evaluate \begin{equation*} \frac{(11^{1} + 1)(11^{2} + 1)(11^{4} + 1)(11^{8} + 1)(11^{16} + 1)(11^{32} + 1)}{(11^{64}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{64}-1)=(11^{32}-1)(11^{32}+1) \end{equation} Cancel $(11^{32}+1)$ both in the numerator and the denominator. \begin{equation} (11^{32}-1)=(11^{16}-1)(11^{16}+1) \end{equation} Cancel $(11^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{64}-1)=(11^{32}-1)(11^{32}+1) \end{equation} Cancel $(11^{32}+1)$ both in the numerator and the denominator. \begin{equation} (11^{32}-1)=(11^{16}-1)(11^{16}+1) \end{equation} Cancel $(11^{16}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Evaluate \begin{equation*} \frac{(11^{1} + 1)(11^{2} + 1)(11^{4} + 1)(11^{8} + 1)(11^{16} + 1)(11^{32} + 1)(11^{64} + 1)}{(11^{128}-1)} \end{equation*}
Select An Answer
Sorry. Please check the correct answer below.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{128}-1)=(11^{64}-1)(11^{64}+1) \end{equation} Cancel $(11^{64}+1)$ both in the numerator and the denominator. \begin{equation} (11^{64}-1)=(11^{32}-1)(11^{32}+1) \end{equation} Cancel $(11^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
Yay! Your are right.
Use $a^2-b^2=(a-b)(a+b)$ \begin{align*} \end{align*} Consider the denominator: \begin{equation} (11^{128}-1)=(11^{64}-1)(11^{64}+1) \end{equation} Cancel $(11^{64}+1)$ both in the numerator and the denominator. \begin{equation} (11^{64}-1)=(11^{32}-1)(11^{32}+1) \end{equation} Cancel $(11^{32}+1)$ both in the numerator and the denominator. Carry on doing it until we are left with $\dfrac{1}{({11}-1)}=\frac{1}{10}$.
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