Solve sqrt{6*2+sqrt[3]{9^2-sqrt{256}-sqrt{1}}}

$$\sqrt { 6 \cdot 2 + \sqrt[ 3 ] { 9 ^ { 2 } - \sqrt { 256 } - \sqrt { 1 } } }$$

Evaluate

$$4$$

Solution Steps

Multiply 6 and 2 to get 12.

$$\sqrt{12+\sqrt[3]{9^{2}-\sqrt{256}-\sqrt{1}}}$$

Calculate 9 to the power of 2 and get 81.

$$\sqrt{12+\sqrt[3]{81-\sqrt{256}-\sqrt{1}}}$$

Calculate the square root of 256 and get 16.

$$\sqrt{12+\sqrt[3]{81-16-\sqrt{1}}}$$

Subtract 16 from 81 to get 65.

$$\sqrt{12+\sqrt[3]{65-\sqrt{1}}}$$

Calculate the square root of 1 and get 1.

$$\sqrt{12+\sqrt[3]{65-1}}$$

Subtract 1 from 65 to get 64.

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

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

$$\sqrt{12+4}$$

Add 12 and 4 to get 16.

$$\sqrt{16}$$

Calculate the square root of 16 and get 4.

$$4$$

Factor

$$2^{2}$$