Answer
$\dfrac{6(x^2-x+1)}{x(2x-1)}, x\ne-1, 0, \frac{1}{2}$
Work Step by Step
Note that:
$\dfrac{12}{x^2+x}\cdot \dfrac{x^3+1}{4x-2}=\dfrac{12}{x^2+x}\cdot \dfrac{x^3+1^3}{4x-2}$
Factor each polynomial completely.
Use special formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and determine the restrictions to obtain:
$=\dfrac{12}{x(x+1)}\cdot \dfrac{(x+1)(x^2-x+1)}{2(2x-1)}, x\ne-1, 0, \frac{1}{2}$
Cancel the common factors $x+1 $ and $2$:
$=\dfrac{6(x^2-x+1)}{x(2x-1)}, x\ne-1, 0, \frac{1}{2}$
Hence, the lowest term is:
$\dfrac{6(x^2-x+1)}{x(2x-1)}, x\ne-1, 0, \frac{1}{2}$
