Solve 6^2*2*(2+2*sqrt{27})

$$6 ^ { 2 } \cdot 2 \cdot ( 2 + 2 \cdot \sqrt { 27 } )$$

Evaluate

$$432\sqrt{3}+144\approx 892.24594887$$

Solution Steps

Calculate 6 to the power of 2 and get 36.

$$2\times 36\left(2+2\sqrt{27}\right)$$

Multiply 2 and 36 to get 72.

$$72\left(2+2\sqrt{27}\right)$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$72\left(2+2\times 3\sqrt{3}\right)$$

Multiply 2 and 3 to get 6.

$$72\left(2+6\sqrt{3}\right)$$

Use the distributive property to multiply 72 by 2+6\sqrt{3}.

$$144+432\sqrt{3}$$

Factor

$$144 {(3 \sqrt{3} + 1)} = 892.24594887$$