Chapter 11.3, Problem 55PPE

To determine

To simplify $\frac{x^{16}-1}{x-1}$ .

Explanation of Solution

Given:

The expression, $\frac{x^{16}-1}{x-1}$

Calculation:

In order to divide the expressions, use the long division method as shown below

  $x-1\begin{array}{c}\hfill x^{15}+x^{14}\mathrm{....}\\ \hfill \overline{)\begin{array}{l}{x}^{16}-1\\ \underset{\_}{-\left(x^{16}-x^{15}\right)}\\ x^{15}-1\\ \underset{\_}{-\left(x^{15}-x^{14}\right)}\\ x^{14}-1\\ \mathrm{......}\end{array}}\end{array}$

Proceeding in the same way it gives the quotient as $x^{15}+x^{14}+x^{13}+\mathrm{...}+x^2+x+1$ .

While on factoring using the difference of squares formula $a^2-b^2=\left(a+b\right)\left(a-b\right)$ , it gives

  $\begin{array}{c}\frac{x^{16}-1}{x-1}=\frac{{\left(x^8\right)}^2-{\left(1\right)}^2}{x-1}\\ =\frac{\left(x^8+1\right)\left(x^8-1\right)}{x-1}\\ =\frac{\left(x^8+1\right)\left({\left(x^4\right)}^2-{\left(1\right)}^2\right)}{x-1}\\ =\frac{\left(x^8+1\right)\left(x^4+1\right)\left(x^4-1\right)}{x-1}\\ =\frac{\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x^2-1\right)}{x-1}\\ =\frac{\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{x-1}\\ =\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x+1\right)\end{array}$

It is easier to use the factoring method as it is not so lengthy and time taking.