Solve sqrt{frac{sqrt[3]{64}+sqrt[4]{256}}{sqrt{64}+sqrt{256}}}

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

Evaluate

$$\frac{\sqrt{3}}{3}\approx 0.577350269$$

Solution Steps

Calculate \sqrt[3]{64} and get 4.

$$\sqrt{\frac{4+\sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$$

Calculate \sqrt[4]{256} and get 4.

$$\sqrt{\frac{4+4}{\sqrt{64}+\sqrt{256}}}$$

Add 4 and 4 to get 8.

$$\sqrt{\frac{8}{\sqrt{64}+\sqrt{256}}}$$

Calculate the square root of 64 and get 8.

$$\sqrt{\frac{8}{8+\sqrt{256}}}$$

Calculate the square root of 256 and get 16.

$$\sqrt{\frac{8}{8+16}}$$

Add 8 and 16 to get 24.

$$\sqrt{\frac{8}{24}}$$

Reduce the fraction \frac{8}{24} to lowest terms by extracting and canceling out 8.

$$\sqrt{\frac{1}{3}}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$\frac{\sqrt{1}}{\sqrt{3}}$$

Calculate the square root of 1 and get 1.

$$\frac{1}{\sqrt{3}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{3}}{3}$$