Answer
$\dfrac{x(x - 1)^3}{x + 4}$
Restrictions: $x \ne -4, 0, 1$
Work Step by Step
Rewrite the exercise using the division ($\div$) symbol:
$(x^2 - x)^2 \div \left[{x(x - 1)^{-2}(x^2 + 3x - 4)}\right]$
Move the expression with the negative exponent from the denominator to the numerator:
$=(x^2 - x)^2(x - 1)^2 \div \left[{x(x^2 + 3x - 4)}\right]$
Invert the divisor and change the operation to multiplication:
$=(x^2 - x)^2(x - 1)^2 \cdot \dfrac{1}{x(x^2 + 3x - 4)}$
Rewrite as one rational expression:
$=\dfrac{(x^2 - x)^2(x - 1)^2}{x(x^2 + 3x - 4)}$
Expand binomials that are raised to a power:
$=\dfrac{(x^2 - x)(x^2 - x)(x - 1)(x - 1)}{x(x^2 + 3x - 4)}$
Factor each polynomial completely:
$=\dfrac{x(x - 1)x(x - 1)(x - 1)(x - 1)}{x(x + 4)(x - 1)}$
Cancel common factors in the numerator and denominator:
$=\dfrac{x(x - 1)^3}{x + 4}$
Restrictions occur when the expression becomes undefined, meaning when the denominator is $0$.
To find the restrictions, set each expression in the denominators of the original rational expressions as well as in the reciprocal equal to zero and solve:
Restriction:
$x = 0$
Restriction:
$x + 4 = 0$
$x = -4$
Restriction:
$x - 1 = 0$
$x = 1$
Therefore, the restrictions are $x \ne -4, 0, 1$.
