Simplify 14/( cube root of 4+5)

$\frac{14}{\sqrt[3]{4} + 5}$

Step 1

Multiply $\frac{14}{\sqrt[3]{4} + 5}$ by $\frac{{\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2}}{{\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2}}$ .

$\frac{14}{\sqrt[3]{4} + 5} \cdot \frac{{\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2}}{{\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2}}$

Step 2

Combine fractions.

Multiply $\frac{14}{\sqrt[3]{4} + 5}$ by $\frac{{\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2}}{{\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2}}$ .

$\frac{14 ({\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2})}{(\sqrt[3]{4} + 5) ({\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2})}$

Expand the denominator using the FOIL method.

$\frac{14 ({\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2})}{{\sqrt[3]{4}}^{3} - 5 {\sqrt[3]{4}}^{2} + \sqrt[3]{4} \cdot 5^{2} + 5 {\sqrt[3]{4}}^{2} - 25 \sqrt[3]{4} + 5^{3}}$

Simplify.

$\frac{14 ({\sqrt[3]{4}}^{2} - \sqrt[3]{4} \cdot 5 + 5^{2})}{129}$

Step 3

Simplify the numerator.

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

$\frac{14 (\sqrt[3]{4^{2}} - \sqrt[3]{4} \cdot 5 + 5^{2})}{129}$

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

$\frac{14 (\sqrt[3]{16} - \sqrt[3]{4} \cdot 5 + 5^{2})}{129}$

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

Factor $8$ out of $16$ .

$\frac{14 (\sqrt[3]{8 (2)} - \sqrt[3]{4} \cdot 5 + 5^{2})}{129}$

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

$\frac{14 (\sqrt[3]{2^{3} \cdot 2} - \sqrt[3]{4} \cdot 5 + 5^{2})}{129}$

Pull terms out from under the radical.

$\frac{14 (2 \sqrt[3]{2} - \sqrt[3]{4} \cdot 5 + 5^{2})}{129}$

Multiply $5$ by $- 1$ .

$\frac{14 (2 \sqrt[3]{2} - 5 \sqrt[3]{4} + 5^{2})}{129}$

Raise $5$ to the power of $2$ .

$\frac{14 (2 \sqrt[3]{2} - 5 \sqrt[3]{4} + 25)}{129}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{14 (2 \sqrt[3]{2} - 5 \sqrt[3]{4} + 25)}{129}$

Decimal Form:

$2.12526911 \dots$

Simplify 14/( cube root of 4+5)

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