Question

Multiply the following and write your answer in lowest terms:

$\frac{x^2-16}{x^2-3x+2}\times \frac{x^2-4}{x^3+64}\times \frac{x^2-4x+16}{x^2-2x-8}$

• A. $\frac{1}{x-1}$
• B. $\frac{1}{x+1}$
• C. $\frac{2}{x-1}$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{x-1}$

The given expression $\frac{x^2-16}{x^2-3x+2}\times \frac{x^2-4}{x^3+64}\times \frac{x^2-4x+16}{x^2-2x-8}$can be simplified as shown below:

$=\frac{x^2-16}{x^2-3x+2}\times \frac{x^2-4}{x^3+64}\times \frac{x^2-4x+16}{x^2-2x-8}$

$=\frac{x^2-4^2}{x^2-x-2x+2}\times \frac{x^2-2^2}{x^3+4^3}\times \frac{x^2-4x+16}{x^2-4x+2x-8}$

$=\frac{(x-4)(x+4)}{x(x-1)-2(x-1)}\times \frac{(x-2)(x+2)}{(x+4)(x^2+4^2-(x\times 4\left)\right)}\times \frac{x^2-4x+16}{x(x-4)+2(x-4)}$

$(\because a^2-b^2=(a-b\left)\right(a+b),a^3+b^3=(a+b\left)\right(a^2+b^2-ab\left)\right)$

$=\frac{(x-4)(x+4)}{(x-1)(x-2)}\times \frac{(x-2)(x+2)}{(x+4)(x^2+16-4x)}\times \frac{x^2+16-4x}{(x-4)(x+2)}$

$=\frac{(x-4)(x+4)(x-2)(x+2)(x^2+16-4x)}{(x-1)(x-2)(x+4)(x^2+16-4x)(x-4)(x+2)}$

$=\frac{1}{x-1}$

Hence, $\frac{x^2-16}{x^2-3x+2}\times \frac{x^2-4}{x^3+64}\times \frac{x^2-4x+16}{x^2-2x-8}=\frac{1}{x-1}$.