Solve 8*(2sqrt{3}+sqrt{6})(2sqrt{3}-6)

$$8 \cdot ( 2 \sqrt { 3 } + \sqrt { 6 } ) ( 2 \sqrt { 3 } - 6 )$$

Evaluate

$$48\left(\sqrt{2}+2-\sqrt{6}-2\sqrt{3}\right)\approx -119.970134186$$

Solution Steps

Use the distributive property to multiply 8 by 2\sqrt{3}+\sqrt{6}.

$$\left(16\sqrt{3}+8\sqrt{6}\right)\left(2\sqrt{3}-6\right)$$

Apply the distributive property by multiplying each term of 16\sqrt{3}+8\sqrt{6} by each term of 2\sqrt{3}-6.

$$32\left(\sqrt{3}\right)^{2}-96\sqrt{3}+16\sqrt{3}\sqrt{6}-48\sqrt{6}$$

The square of \sqrt{3} is 3.

$$32\times 3-96\sqrt{3}+16\sqrt{3}\sqrt{6}-48\sqrt{6}$$

Multiply 32 and 3 to get 96.

$$96-96\sqrt{3}+16\sqrt{3}\sqrt{6}-48\sqrt{6}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$96-96\sqrt{3}+16\sqrt{3}\sqrt{3}\sqrt{2}-48\sqrt{6}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$96-96\sqrt{3}+16\times 3\sqrt{2}-48\sqrt{6}$$

Multiply 16 and 3 to get 48.

$$96-96\sqrt{3}+48\sqrt{2}-48\sqrt{6}$$

Factor

$$48 {(\sqrt{2} + 2 - \sqrt{6} - 2 \sqrt{3})} = -119.970134186$$