Question 276413
{{{root(3, 32x^6y^9/z^8)}}}
Simplifying a cube root involves finding perfect cubes. Since there is a denominator we will also want to rationalize the denominator. I'm going to make the denominator a perfect cube at the start. This will take care of the rationalizing denomiinator before we start. Since any exponent that is divisible by 3 is a perfect cube, all I need to do is change {{{z^8}}} to {{{z^9}}}:
{{{root(3, (32x^6y^9/z^8)(z/z))}}}
which simplifies to:
{{{root(3, 32x^6y^9z/z^9)}}}
Now we can start finding perfect cubes:
{{{root(3, 2^3*4(x^2)^3(y^3)^3z/(z^3)^3)}}}
Using the properties of radicals we can separate out all the perfect cubes:
{{{(root(3, 2^3)*root(3, 4)*root(3, (x^2)^3)*root(3, (y^3)^3)*root(3, z))/root(3, (z^3)^3))}}}
Replacing the cube roots of the perfect cubes we get:
{{{(2*root(3, 4)*x^2*y^3*root(3, z))/z^3}}}
Rearranging the factors in the numerator and combining the remaining radicals we get:
{{{(2x^2y^3*root(3, 4z))/z^3}}}