Simplify cube root of (9x^6)/(y^3z^9)

$\sqrt[3]{\frac{9 x^{6}}{y^{3} z^{9}}}$

Rewrite $\frac{9 x^{6}}{y^{3} z^{9}}$ as ${(\frac{x^{2}}{y z^{3}})}^{3} \cdot 9$ .

Factor the perfect power ${(x^{2})}^{3}$ out of $9 x^{6}$ .

$\sqrt[3]{\frac{{(x^{2})}^{3} \cdot 9}{y^{3} z^{9}}}$

Factor the perfect power ${(y z^{3})}^{3}$ out of $y^{3} z^{9}$ .

$\sqrt[3]{\frac{{(x^{2})}^{3} \cdot 9}{{(y z^{3})}^{3} \cdot 1}}$

Rearrange the fraction $\frac{{(x^{2})}^{3} \cdot 9}{{(y z^{3})}^{3} \cdot 1}$ .

$\sqrt[3]{{(\frac{x^{2}}{y z^{3}})}^{3} \cdot 9}$

Pull terms out from under the radical.

$\frac{x^{2}}{y z^{3}} \sqrt[3]{9}$

Combine $\frac{x^{2}}{y z^{3}}$ and $\sqrt[3]{9}$ .

$\frac{x^{2} \sqrt[3]{9}}{y z^{3}}$