Verified answer

24 = 2^3*3

x^5 = x^3*x^2

y^7 = y^3*y^3*y

z^9 = z^3*z^3*z^3

Multiply all the perfect cubes outside the radical and leave the non-perfect cubes inside. Eliminate the 3 as an exponent on the outside terms.

(2*x*y*y*z*z*z)* ^3√(3*x^2*y)

2xy^2z^3 ^3√(3x^2y)

I take it the whole thing is in the cuberoot

So in a cube root, "it takes 3 to get out of jail"

2*2*2*3 = 24

And then we have 5 x's so 3 come out, 7y's so 6 come out and 9 z's so they all come out.

For every 3 of a number or variable we have, it comes out in the form of 1.

So we have:

2xy^2z^3 ^3√3x^2y