# How do you simplify (-2x+root3(-5x^4y^3))/(3root3(15x^3y))?

Jul 9, 2017

$\frac{- 2 - y \sqrt[3]{5 x}}{3 \sqrt[3]{15 y}}$

#### Explanation:

First, pull any cubes out of the cube roots. We know that:

$\sqrt[3]{{\textcolor{red}{a}}^{3}} = \textcolor{red}{a}$

And therefore,

root(3)(color(red)a^3 xx b xx c xx cdots) = color(red)a * root(3)(b xx c xx cdots

We can use this to pull stuff out of the cube roots which doesn't need to be in there. So let's do that $-$ I'll highlight any cubes with different colors so it's clearer what I'm doing when I change things around.

$\frac{- 2 x + \sqrt[3]{- 5 {x}^{4} {y}^{3}}}{3 \sqrt[3]{15 {x}^{3} y}}$

$= \frac{- 2 x + \sqrt[3]{5 \left(- 1\right) \left({x}^{3}\right) \left(x\right) \left({y}^{3}\right)}}{3 \sqrt[3]{15 \left({x}^{3}\right) \left(y\right)}}$

$= \frac{- 2 x + \sqrt[3]{5 {\left(\textcolor{red}{- 1}\right)}^{3} \left({\textcolor{b l u e}{x}}^{3}\right) \left(x\right) \left({\textcolor{\mathmr{and} a n \ge}{y}}^{3}\right)}}{3 \sqrt[3]{15 \left({\textcolor{\lim e g r e e n}{x}}^{3}\right) \left(y\right)}}$

$= \frac{- 2 x + \left(\textcolor{red}{- 1}\right) \left(\textcolor{b l u e}{x}\right) \left(\textcolor{\mathmr{and} a n \ge}{y}\right) \sqrt[3]{5 x}}{3 \left(\textcolor{\lim e g r e e n}{x}\right) \sqrt[3]{15 \left(y\right)}}$

$= \frac{- 2 x - x y \sqrt[3]{5 x}}{3 x \sqrt[3]{15 y}}$

At this point, we can cancel out an $x$ on the top and the bottom. Notice that all terms have an $x$ outside the radical, so this is possible.

$= \frac{\left(x\right) \left(- 2 - y \sqrt[3]{5 x}\right)}{\left(x\right) \left(3 \sqrt[3]{15 y}\right)}$

$= \frac{\cancel{\left(x\right)} \left(- 2 - y \sqrt[3]{5 x}\right)}{\cancel{\left(x\right)} \left(3 \sqrt[3]{15 y}\right)}$

$= \frac{- 2 - y \sqrt[3]{5 x}}{3 \sqrt[3]{15 y}}$

We could technically simplify this further by rationalizing the denominator, but this is a pretty good stopping point as far as simplification goes.