1. Simplifying Radicals and Exponents Problem

How would you solve:

((25x^3)^(2/3))) / ((125xy^2)^(1/3)))

?

I got the cube root of 625x^6 in the numerator and the cube root of 125xy^2 in the denominator.
After I simplified them, I got 5x^2 * (cube root of 5) in the numerator and 5(xy^2)^(1/3) in the denominator, but that doesn't look right to me.

2. $\frac{(25x^3)^{2/3}}{(125xy^2)^{1/3}} = \frac{\sqrt[3]{125 \cdot 5}x^2}{5\sqrt[3]{x}y^{2/3}} = \boxed{\sqrt[3]{5}x^{5/3}y^{-2/3}}$

That seems good to me but I'm kind of tired.

3. Hello, MathGeek06!

TrevorP is absolutely correct . . .

Simplify: . $\frac{(25x^3)^{\frac{2}{3}}} {(125xy^2)^{\frac{1}{3}}}$
I would do it like this . . .

. . $\frac{\left(5^2x^3\right)^{\frac{2}{3}}}{\left(5^3 xy^2\right)^{\frac{1}{3}}} \;=\;\frac{(5^2)^{\frac{2}{3}}(x^3)^{\frac{2}{3}}} {(5^3)^{\frac{1}{3}}(x^{\frac{1}{3}})(y^2)^{\frac{ 1}{3}}} \;=\; \frac{5^{\frac{4}{3}}x^2} {5^1x^{\frac{1}{3}}y^{\frac{2}{3}} } \;=\;\boxed{\frac{5^{\frac{1}{3}}x^{\frac{5}{3}}}{ y^{\frac{2}{3}}}}$