Simplify (x^3-9x)/(x^2+6x+9)*(x^3+3x^2)/(x-3)

Factor ${\displaystyle x}$ out of ${\displaystyle x^3-9x}$.

Rewrite ${\displaystyle 9}$ as ${\displaystyle 3^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=3}$.

Rewrite ${\displaystyle 9}$ as ${\displaystyle 3^2}$.

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

Rewrite the polynomial.

Factor using the perfect square trinomial rule ${\displaystyle a^2+2ab+b^2={(a+b)}^2}$, where ${\displaystyle a=x}$ and ${\displaystyle b=3}$.

Factor ${\displaystyle x^2}$ out of ${\displaystyle x^3+3x^2}$.

Cancel the common factor of ${\displaystyle x-3}$.

Cancel the common factor of ${\displaystyle x+3}$.

Combine ${\displaystyle \frac{x(x+3)}{x+3}}$ and ${\displaystyle x^2}$.

Move ${\displaystyle x^2}$.

Multiply ${\displaystyle x^2}$ by ${\displaystyle x}$.

Add ${\displaystyle 2}$ and ${\displaystyle 1}$.

Cancel the common factor.

Divide ${\displaystyle x^3}$ by ${\displaystyle 1}$.