$$\sqrt[3]{\frac{11a^2}{64b^6}}$$

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Aniqa O'Neill

Simplilfying the given radical expression (involving fractional exponents) means removing roots as much as possible, using the laws of exponents (powers)
First apply the rule
$$\sqrt[n]{\frac{x}{y}}=\sqrt[n]{\frac{x}{\sqrt[n]{y}}}$$
Here, $$\displaystyle{x}={11}{a}^{{2}},{y}={64}{b}^{{6}}={\left({4}\right)}^{{3}}{b}^{{6}}={\left({4}{b}^{{2}}\right)}^{{3}}$$
So,
$$\sqrt[3]{\frac{11a^2}{64b^6}}=\sqrt[3]{\frac{{11a^2}}{\sqrt[3]{(4b^2)^3}}}$$
Note: the numerator (radical ) cannot be simplified as none of the factors 11 or a^2 is a cube. On the other hand, the denominator radical can be simplified , as it is a cube.
Now apply the rule
$$\sqrt[n]{z^k}=(z^k)^{\frac{1}{n}}=z^{\frac{k}{n}}$$
Here, $$\displaystyle{z}={4}{b}^{{2}},{k}={3},{n}={3}$$
So,
$$\sqrt[3]{(4b^2)^3}=(4b^2)^{\frac{3}{3}}=4b^2$$
Answer: $$\sqrt[3]{\frac{11a^2}{64b^6}}=\frac{\sqrt[3]{11a^2}}{4b^2}$$