Evaluate cube root of 11/64-( cube root of 11)/9

$\sqrt[3]{\frac{11}{64}} - \frac{\sqrt[3]{11}}{9}$

Step 1

Simplify each term.

Rewrite $\sqrt[3]{\frac{11}{64}}$ as $\frac{\sqrt[3]{11}}{\sqrt[3]{64}}$ .

$\frac{\sqrt[3]{11}}{\sqrt[3]{64}} - \frac{\sqrt[3]{11}}{9}$

Simplify the denominator.

Rewrite $64$ as $4^{3}$ .

$\frac{\sqrt[3]{11}}{\sqrt[3]{4^{3}}} - \frac{\sqrt[3]{11}}{9}$

Pull terms out from under the radical, assuming real numbers.

$\frac{\sqrt[3]{11}}{4} - \frac{\sqrt[3]{11}}{9}$

Step 2

To write $\frac{\sqrt[3]{11}}{4}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\frac{\sqrt[3]{11}}{4} \cdot \frac{9}{9} - \frac{\sqrt[3]{11}}{9}$

Step 3

To write $- \frac{\sqrt[3]{11}}{9}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{\sqrt[3]{11}}{4} \cdot \frac{9}{9} - \frac{\sqrt[3]{11}}{9} \cdot \frac{4}{4}$

Step 4

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{\sqrt[3]{11}}{4}$ by $\frac{9}{9}$ .

$\frac{\sqrt[3]{11} \cdot 9}{4 \cdot 9} - \frac{\sqrt[3]{11}}{9} \cdot \frac{4}{4}$

Multiply $4$ by $9$ .

$\frac{\sqrt[3]{11} \cdot 9}{36} - \frac{\sqrt[3]{11}}{9} \cdot \frac{4}{4}$

Multiply $\frac{\sqrt[3]{11}}{9}$ by $\frac{4}{4}$ .

$\frac{\sqrt[3]{11} \cdot 9}{36} - \frac{\sqrt[3]{11} \cdot 4}{9 \cdot 4}$

Multiply $9$ by $4$ .

$\frac{\sqrt[3]{11} \cdot 9}{36} - \frac{\sqrt[3]{11} \cdot 4}{36}$

Step 5

Combine the numerators over the common denominator.

$\frac{\sqrt[3]{11} \cdot 9 - \sqrt[3]{11} \cdot 4}{36}$

Step 6

Simplify the numerator.

Move $9$ to the left of $\sqrt[3]{11}$ .

$\frac{9 \cdot \sqrt[3]{11} - \sqrt[3]{11} \cdot 4}{36}$

Multiply $4$ by $- 1$ .

$\frac{9 \sqrt[3]{11} - 4 \sqrt[3]{11}}{36}$

Subtract $4 \sqrt[3]{11}$ from $9 \sqrt[3]{11}$ .

$\frac{5 \sqrt[3]{11}}{36}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{5 \sqrt[3]{11}}{36}$

Decimal Form:

$0.30888612 \dots$