# How do you simplify root3(81x^8 y^12)?

${3}^{\frac{4}{3}} {x}^{\frac{8}{3}} {y}^{4} = 3 {x}^{2} {y}^{4} \sqrt[3]{3 {x}^{2}}$

#### Explanation:

With the expression $\sqrt[3]{81 {x}^{8} {y}^{12}}$, we're taking the cube root of all the terms. We can write the cube root as a power of $\frac{1}{3}$, and so we get:

${\left(81 {x}^{8} {y}^{12}\right)}^{\frac{1}{3}}$

We can now distribute the $\frac{1}{3}$ exponent to all the individual terms:

${81}^{\frac{1}{3}} \times {\left({x}^{8}\right)}^{\frac{1}{3}} \times {\left({y}^{12}\right)}^{\frac{1}{3}}$

I'm going to express $81 = {3}^{4}$ and will also use the rule that ${\left({x}^{a}\right)}^{b} = {x}^{a b}$:

${3}^{4 \times \left(\frac{1}{3}\right)} \times {x}^{8 \times \left(\frac{1}{3}\right)} \times {y}^{12 \times \left(\frac{1}{3}\right)}$

which we can simplify:

${3}^{\frac{4}{3}} \times {x}^{\frac{8}{3}} \times {y}^{\frac{12}{3}}$

${3}^{\frac{4}{3}} \times {x}^{\frac{8}{3}} \times {y}^{4}$

We can now take the fractional exponents and where there is even divisibility, that lies outside of the root. Where there is a remainder, that can stay as a fraction or be shown under a root sign:

${3}^{\frac{3}{3}} \times {3}^{\frac{1}{3}} \times {x}^{\frac{6}{3}} \times {x}^{\frac{2}{3}} \times {y}^{4}$

$3 {x}^{2} {y}^{4} {3}^{\frac{1}{3}} {x}^{\frac{2}{3}} = 3 {x}^{2} {y}^{4} \sqrt[3]{3 {x}^{2}}$