Solve 2^3sqrt{2}-3^3sqrt{16}+2^3sqrt{128}

$$2 ^ { 3 } \sqrt { 2 } - 3 ^ { 3 } \sqrt { 16 } + 2 ^ { 3 } \sqrt { 128 }$$

Evaluate

$$36\left(2\sqrt{2}-3\right)\approx -6.176623509$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$8\sqrt{2}-3^{3}\sqrt{16}+2^{3}\sqrt{128}$$

Calculate 3 to the power of 3 and get 27.

$$8\sqrt{2}-27\sqrt{16}+2^{3}\sqrt{128}$$

Calculate the square root of 16 and get 4.

$$8\sqrt{2}-27\times 4+2^{3}\sqrt{128}$$

Multiply 27 and 4 to get 108.

$$8\sqrt{2}-108+2^{3}\sqrt{128}$$

Calculate 2 to the power of 3 and get 8.

$$8\sqrt{2}-108+8\sqrt{128}$$

Factor 128=8^{2}\times 2. Rewrite the square root of the product \sqrt{8^{2}\times 2} as the product of square roots \sqrt{8^{2}}\sqrt{2}. Take the square root of 8^{2}.

$$8\sqrt{2}-108+8\times 8\sqrt{2}$$

Multiply 8 and 8 to get 64.

$$8\sqrt{2}-108+64\sqrt{2}$$

Combine 8\sqrt{2} and 64\sqrt{2} to get 72\sqrt{2}.

$$72\sqrt{2}-108$$