# How do you simplify [(x^3-3x^2)/ (3x+6)] / [(x^3-8x^2+15x) / (6x^2-18x-60)?

Jul 20, 2015

I found: $2 x \frac{x + 2}{x + 3}$

#### Explanation:

I tried factorizing as much as possible to get (changing the division into a multiplication):

$= \frac{{x}^{2} \left(x - 3\right)}{3 \left(x + 3\right)} \times \frac{6 {x}^{2} - 18 x - 60}{x \left({x}^{2} - 8 x + 15\right)} =$

$= \frac{{x}^{2} \left(x - 3\right)}{3 \left(x + 3\right)} \times \frac{6 \left(x + 2\right) \left(x - 5\right)}{x \left(x - 3\right) \left(x - 5\right)} =$

$= \frac{{x}^{\cancel{2}} \cancel{\left(x - 3\right)}}{\cancel{3} \left(x + 3\right)} \times \frac{{\cancel{6}}^{2} \left(x + 2\right) \cancel{\left(x - 5\right)}}{\cancel{x} \cancel{\left(x - 3\right)} \cancel{\left(x - 5\right)}} =$

$= 2 x \frac{x + 2}{x + 3}$