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

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Answer 1 · 2015-07-20T00:51:38

I found: $2x(x+2)/(x+3)$

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

$=(x^2(x-3))/(3(x+3))xx(6x^2-18x-60)/(x(x^2-8x+15))=$

$=(x^2(x-3))/(3(x+3))xx(6(x+2)(x-5))/(x(x-3)(x-5))=$

$=(x^cancel(2)cancel((x-3)))/(cancel(3)(x+3))xx(cancel(6)^2(x+2)cancel((x-5)))/(cancel(x)cancel((x-3))cancel((x-5)))=$

$=2x(x+2)/(x+3)$

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