Question 405428
{{{root(3, (8m^7n^9)/(n^2m^2))}}}
Your first step is right on. The radicand (the expression inside the radical) simplifies to:
{{{root(3, 8m^5n^7)}}}
Next we look for factors of the radicand that are perfect cubes. As you already found, 8 is a perfect cube. But there are more perfect cube factors. Because of the way exponents work, the exponent on a perfect cube is <i>not a perfect cube</i> but a multiple of 3! So {{{x^3}}}, {{{y^12}}}, {{{z^300}}} are all perfect cubes, even though 3, 12 and 300 are not themselves perfect cubes.<br>
So your radicand factored into as many perfect cubes as we can find is:
{{{root(3, 8m^3*m^2*n^3*n^3*n)}}}
For reasons that will become clear shortly I like to use the Commutative Property to rearrange the order of the factors so that all the perfect cubes are in front:
{{{root(3, 8m^3*n^3*n^3*m^2*n)}}}
Next we use a property of radicals, {{{root(a, p*q) = root(a, p)*root(3, q)}}}, to split this cube root of a product into a product of cube roots. We want each perfect cube factor in its own cube root. The factors that are not perfect cubes can all go into one cube root:
{{{root(3, 8)*root(3, m^3)*root(3, n^3)*root(3, n^3)*root(3, m^2*n)}}}
The cube roots of the perfect cubes will simplify:
{{{2*m*n*n*root(3, m^2*n)}}}
or
{{{2*m*n^2*root(3, m^2*n)}}}
This is the simplified cube root. (Note how the radical is at the end. This is the usual way to write terms like this and it is the reason I put all the perfect cubes n the front earlier.)