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

1 Answer
Jul 9, 2017

Answer:

#(-2 -yroot(3)(5x))/(3root(3)(15y))#

Explanation:

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

#root(3)(color(red)a^3) = color(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.

#(-2x + root(3)(-5x^4y^3))/(3root(3)(15x^3y))#

#= (-2x + root(3)(5(-1)(x^3)(x)(y^3)))/(3root(3)(15(x^3)(y)))#

#= (-2x + root(3)(5(color(red)(-1))^3(color(blue)x^3)(x)(color(orange)y^3)))/(3root(3)(15(color(limegreen)x^3)(y)))#

#= (-2x + (color(red)(-1))(color(blue)x)(color(orange)y) root(3)(5x))/(3(color(limegreen)x)root(3)(15(y)))#

#= (-2x -xyroot(3)(5x))/(3xroot(3)(15y))#

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.

#= ((x)(-2 -yroot(3)(5x)))/((x)(3root(3)(15y)))#

#= (cancel((x))(-2 -yroot(3)(5x)))/(cancel((x))(3root(3)(15y)))#

#= (-2 -yroot(3)(5x))/(3root(3)(15y))#

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

Final Answer