Question

(a) Verify that $x-1$is a factor of $x_n-1$.

(b) If n is a positive integer, prove that the prime factorization of role="math" localid="1646128650215" $2^{2n}·3^n-1$includes 11 as one of the prime factors.

Short answer

1. It is proved that $x-1$is the factor of $x^n-1$.
2. It is proved that$2^{2n}·3^n-1$ includes 11 as one of the prime factors.

Step-by-step solution

Prove that x-1 is a factor of xn-1.

Rewrite the given expression$x^n-1$ as follows:

$x^n-1=\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+x+1\right)$

The expression$x^n-1$ can be written as the multiple of $x-1$. This implies that $x-1$is the factor of $x^n-1$.

Prove that 22n·3n-1 includes 11 as one of the prime factors.

Rewrite the given expression as follows:$\begin{array}{rcl}{2}^{2n}·3^n-1& =& {\left(2^2·3\right)}^n-1\\ & =& {\left(4·3\right)}^n-1\\ & =& {12}^n-1\end{array}$

From step-1, $x-1$is the factor of$x^n-1$ . This implies that$12-1=11$ is the factor of ${12}^n-1$.

This implies$2^{2n}·3^n-1$ includes 11 as one of the prime factors.

Hence, it is proved that$2^{2n}·3^n-1$ includes 11 as one of the prime factors.