Answer
$\dfrac{x^{3}-6x^{2}+9x}{x^{2}+4x-21}=\dfrac{x(x-3)}{x+7}$
Work Step by Step
$\dfrac{x^{3}-6x^{2}+9x}{x^{2}+4x-21}$
Take out common factor $x$ from the numerator and factor the denominator:
$\dfrac{x^{3}-6x^{2}+9x}{x^{2}+4x-21}=\dfrac{x(x^{2}-6x+9)}{(x+7)(x-3)}=...$
Factor the parentheses in the numerator and simplify:
$...=\dfrac{x(x-3)^{2}}{(x+7)(x-3)}=\dfrac{x(x-3)}{x+7}$
