Question

Simplify the radicals. Assume that all variables can be any real number.

Answer 1

$\sqrt[3]{\frac{-27x^9}{64y^{12}}}$

we take cube root separately

$\sqrt[3]{-27} =\sqrt[3]{-3*-3*-3} =-3$

We know cuberoot (x^3) is x

$\sqrt[3]{x^9}=\sqrt[3]{x^3*x^3*x^3}=x*x*x= x^3$

$\sqrt[3]{64} =\sqrt[3]{4*4*4}= 4$

$\sqrt[3]{y^{12}} =\sqrt[3]{y^3*y^3*y^3*y^3}= y*y*y*y= y^4$

$\sqrt[3]{\frac{-27x^9}{64y^{12}}}=\frac{-3x^3}{4y^4}$