Simplify cube root of 7/16

$\sqrt[3]{\frac{7}{16}}$

Step 1

Rewrite $\sqrt[3]{\frac{7}{16}}$ as $\frac{\sqrt[3]{7}}{\sqrt[3]{16}}$ .

$\frac{\sqrt[3]{7}}{\sqrt[3]{16}}$

Step 2

Simplify the denominator.

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

Factor $8$ out of $16$ .

$\frac{\sqrt[3]{7}}{\sqrt[3]{8 (2)}}$

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

$\frac{\sqrt[3]{7}}{\sqrt[3]{2^{3} \cdot 2}}$

Pull terms out from under the radical.

$\frac{\sqrt[3]{7}}{2 \sqrt[3]{2}}$

Step 3

Multiply $\frac{\sqrt[3]{7}}{2 \sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$\frac{\sqrt[3]{7}}{2 \sqrt[3]{2}} \cdot \frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$

Step 4

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{7}}{2 \sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 \sqrt[3]{2} {\sqrt[3]{2}}^{2}}$

Move $\sqrt[3]{2}$ .

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 (\sqrt[3]{2} {\sqrt[3]{2}}^{2})}$

Raise $\sqrt[3]{2}$ to the power of $1$ .

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 ({\sqrt[3]{2}}^{1} {\sqrt[3]{2}}^{2})}$

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

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 {\sqrt[3]{2}}^{1 + 2}}$

Add $1$ and $2$ .

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 {\sqrt[3]{2}}^{3}}$

Rewrite ${\sqrt[3]{2}}^{3}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2}$ as $2^{\frac{1}{3}}$ .

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 {(2^{\frac{1}{3}})}^{3}}$

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

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 \cdot 2^{\frac{1}{3} \cdot 3}}$

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

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 \cdot 2^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 \cdot 2^{\frac{3}{3}}}$

Rewrite the expression.

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 \cdot 2^{1}}$

Evaluate the exponent.

$\frac{\sqrt[3]{7} {\sqrt[3]{2}}^{2}}{2 \cdot 2}$

Step 5

Simplify the numerator.

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

$\frac{\sqrt[3]{7} \sqrt[3]{2^{2}}}{2 \cdot 2}$

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

$\frac{\sqrt[3]{7} \sqrt[3]{4}}{2 \cdot 2}$

Step 6

Multiply $2$ by $2$ .

$\frac{\sqrt[3]{7} \sqrt[3]{4}}{4}$

Step 7

Simplify the numerator.

Combine using the product rule for radicals.

$\frac{\sqrt[3]{7 \cdot 4}}{4}$

Multiply $7$ by $4$ .

$\frac{\sqrt[3]{28}}{4}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt[3]{28}}{4}$

Decimal Form:

$0.75914724 \dots$

Simplify cube root of 7/16

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