Simplify 3 fourth root of 256- cube root of 64+ square root of 169

$3 \sqrt[4]{256} - \sqrt[3]{64} + \sqrt{169}$

Step 1

Simplify each term.

Rewrite $256$ as $4^{4}$ .

$3 \sqrt[4]{4^{4}} - \sqrt[3]{64} + \sqrt{169}$

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

$3 \cdot 4 - \sqrt[3]{64} + \sqrt{169}$

Multiply $3$ by $4$ .

$12 - \sqrt[3]{64} + \sqrt{169}$

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

$12 - \sqrt[3]{4^{3}} + \sqrt{169}$

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

$12 - 1 \cdot 4 + \sqrt{169}$

Multiply $- 1$ by $4$ .

$12 - 4 + \sqrt{169}$

Rewrite $169$ as $13^{2}$ .

$12 - 4 + \sqrt{13^{2}}$

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

$12 - 4 + 13$

Step 2

Simplify by adding and subtracting.

Subtract $4$ from $12$ .

$8 + 13$

Add $8$ and $13$ .

$21$