Simplify cube root of (432n^12)/(64q^6)

$\sqrt[3]{\frac{432 n^{12}}{64 q^{6}}}$

Reduce the expression $\frac{432 n^{12}}{64 q^{6}}$ by cancelling the common factors.

Factor $16$ out of $432 n^{12}$ .

$\sqrt[3]{\frac{16 (27 n^{12})}{64 q^{6}}}$

Factor $16$ out of $64 q^{6}$ .

$\sqrt[3]{\frac{16 (27 n^{12})}{16 (4 q^{6})}}$

Cancel the common factor.

$\sqrt[3]{\frac{16 (27 n^{12})}{16 (4 q^{6})}}$

Rewrite the expression.

$\sqrt[3]{\frac{27 n^{12}}{4 q^{6}}}$

Rewrite $\frac{27 n^{12}}{4 q^{6}}$ as ${(\frac{3 n^{4}}{q^{2}})}^{3} \frac{1}{4}$ .

Factor the perfect power ${(3 n^{4})}^{3}$ out of $27 n^{12}$ .

$\sqrt[3]{\frac{{(3 n^{4})}^{3} \cdot 1}{4 q^{6}}}$

Factor the perfect power ${(q^{2})}^{3}$ out of $4 q^{6}$ .

$\sqrt[3]{\frac{{(3 n^{4})}^{3} \cdot 1}{{(q^{2})}^{3} \cdot 4}}$

Rearrange the fraction $\frac{{(3 n^{4})}^{3} \cdot 1}{{(q^{2})}^{3} \cdot 4}$ .

$\sqrt[3]{{(\frac{3 n^{4}}{q^{2}})}^{3} \frac{1}{4}}$

Pull terms out from under the radical.

$\frac{3 n^{4}}{q^{2}} \sqrt[3]{\frac{1}{4}}$

Rewrite $\sqrt[3]{\frac{1}{4}}$ as $\frac{\sqrt[3]{1}}{\sqrt[3]{4}}$ .

$\frac{3 n^{4}}{q^{2}} \cdot \frac{\sqrt[3]{1}}{\sqrt[3]{4}}$

Combine.

$\frac{3 n^{4} \sqrt[3]{1}}{q^{2} \sqrt[3]{4}}$

Simplify the numerator.

Any root of $1$ is $1$ .

$\frac{3 n^{4} \cdot 1}{q^{2} \sqrt[3]{4}}$

Multiply $3$ by $1$ .

$\frac{3 n^{4}}{q^{2} \sqrt[3]{4}}$

Multiply $\frac{3 n^{4}}{q^{2} \sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

$\frac{3 n^{4}}{q^{2} \sqrt[3]{4}} \cdot \frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{3 n^{4}}{q^{2} \sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} \sqrt[3]{4} {\sqrt[3]{4}}^{2}}$

Move $\sqrt[3]{4}$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} (\sqrt[3]{4} {\sqrt[3]{4}}^{2})}$

Raise $\sqrt[3]{4}$ to the power of $1$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} ({\sqrt[3]{4}}^{1} {\sqrt[3]{4}}^{2})}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} {\sqrt[3]{4}}^{1 + 2}}$

Add $1$ and $2$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} {\sqrt[3]{4}}^{3}}$

Rewrite ${\sqrt[3]{4}}^{3}$ as $4$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4}$ as $4^{\frac{1}{3}}$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} {(4^{\frac{1}{3}})}^{3}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} \cdot 4^{\frac{1}{3} \cdot 3}}$

Combine $\frac{1}{3}$ and $3$ .

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} \cdot 4^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} \cdot 4^{\frac{3}{3}}}$

Rewrite the expression.

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} \cdot 4^{1}}$

Evaluate the exponent.

$\frac{3 n^{4} {\sqrt[3]{4}}^{2}}{q^{2} \cdot 4}$

Simplify the numerator.

Rewrite ${\sqrt[3]{4}}^{2}$ as ${(4^{2})}^{\frac{1}{3}}$ .

$\frac{3 n^{4} \sqrt[3]{4^{2}}}{q^{2} \cdot 4}$

Raise $4$ to the power of $2$ .

$\frac{3 n^{4} \sqrt[3]{16}}{q^{2} \cdot 4}$

Rewrite $16$ as $2^{3} \cdot 2$ .

Factor $8$ out of $16$ .

$\frac{3 n^{4} \sqrt[3]{8 (2)}}{q^{2} \cdot 4}$

Rewrite $8$ as $2^{3}$ .

$\frac{3 n^{4} \sqrt[3]{2^{3} \cdot 2}}{q^{2} \cdot 4}$

Pull terms out from under the radical.

$\frac{3 n^{4} \cdot 2 \sqrt[3]{2}}{q^{2} \cdot 4}$

Multiply $2$ by $3$ .

$\frac{6 n^{4} \sqrt[3]{2}}{q^{2} \cdot 4}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $6$ and $4$ .

Factor $2$ out of $6 n^{4} \sqrt[3]{2}$ .

$\frac{2 (3 n^{4} \sqrt[3]{2})}{q^{2} \cdot 4}$

Cancel the common factors.

Factor $2$ out of $q^{2} \cdot 4$ .

$\frac{2 (3 n^{4} \sqrt[3]{2})}{2 (q^{2} \cdot 2)}$

Cancel the common factor.

$\frac{2 (3 n^{4} \sqrt[3]{2})}{2 (q^{2} \cdot 2)}$

Rewrite the expression.

$\frac{3 n^{4} \sqrt[3]{2}}{q^{2} \cdot 2}$

Move $2$ to the left of $q^{2}$ .

$\frac{3 n^{4} \sqrt[3]{2}}{2 q^{2}}$

Simplify cube root of (432n^12)/(64q^6)

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