Answer
$(x-1)(2x-1)$.
Work Step by Step
The given expression is
$=\frac{8x^3-1}{4x^2+2x+1}\div \frac{x-1}{(x-1)^2}$
$=\frac{8x^3-1}{4x^2+2x+1}\cdot \frac{(x-1)^2}{x-1}$
Factor each denominator and numerator.
$=8x^3-1$
$=(2x)^3-1^3$
Use the algebraic identity $a^3-b^3=(a-b)(a^2+ab+b^2)$.
$=(2x-1)(4x^2+2x+1)$
$=4x^2+2x+1$
This is a prime polynomial.
$=x-1$
This is a prime polynomial.
$=(x-1)^2$
$=(x-1)(x-1)$.
Plug all factors into the given expression.
$=\frac{(2x-1)(4x^2+2x+1)}{4x^2+2x+1}\cdot \frac{(x-1)(x-1)}{x-1}$
Cancel common terms.
$=(2x-1)(x-1)$
Rearrange.
$=(x-1)(2x-1)$.
