Answer
$\dfrac{x^{2}+x-42}{x-3}\cdot\dfrac{(x-3)^{2}}{x+7}=(x-3)(x+6)$
Work Step by Step
$\dfrac{x^{2}+x-42}{x-3}\cdot\dfrac{(x-3)^{2}}{x+7}$
Factor the numerator of the first fraction:
$\dfrac{x^{2}+x-42}{x-3}\cdot\dfrac{(x-3)^{2}}{x+7}=\dfrac{(x+7)(x-6)}{x-3}\cdot\dfrac{(x-3)^{2}}{x+7}=...$
Evaluate the product and simplify by removing the factors that appear both in numerator and the denominator of the resulting expression:
$...=\dfrac{(x-3)^{2}(x+7)(x-6)}{(x-3)(x+7)}=(x-3)(x-6)$
