Answer
$\dfrac{6(x^2-x+1)}{x(2x-1)};x \ne 0,-1,\frac{1}{2}$
Work Step by Step
Factor each polynomial completely to obtain:
$=\dfrac{2(2)(3)}{x(x+1)} \cdot \dfrac{(x+1)(x^2-x+1)}{2(2x-1)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{2\cancel{(2)}(3)}{x\cancel{(x+1)}} \cdot \dfrac{\cancel{(x+1)}(x^2-x+1)}{\cancel{2}(2x-1)}
\\=\dfrac{2(3)(x^2-x+1)}{x(2x-1)}
\\=\dfrac{6(x^2-x+1)}{x(2x-1)};x \ne 0,-1,\frac{1}{2}$
($x$ cannot be $-1$, $0$, and $\dfrac{1}{2}$ because they make the denominator equal to zero, which in turn makes the rational expression/s undefined.)
