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

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

Evaluate

$$6\left(\sqrt{3}-1\right)\approx 4.392304845$$

Solution Steps

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

$$6\sqrt{6}\sqrt{2}+2\left(\sqrt{6}\right)^{2}-9\left(\sqrt{2}\right)^{2}-3\sqrt{2}\sqrt{6}$$

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

$$6\sqrt{2}\sqrt{3}\sqrt{2}+2\left(\sqrt{6}\right)^{2}-9\left(\sqrt{2}\right)^{2}-3\sqrt{2}\sqrt{6}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$6\times 2\sqrt{3}+2\left(\sqrt{6}\right)^{2}-9\left(\sqrt{2}\right)^{2}-3\sqrt{2}\sqrt{6}$$

Multiply 6 and 2 to get 12.

$$12\sqrt{3}+2\left(\sqrt{6}\right)^{2}-9\left(\sqrt{2}\right)^{2}-3\sqrt{2}\sqrt{6}$$

The square of \sqrt{6} is 6.

$$12\sqrt{3}+2\times 6-9\left(\sqrt{2}\right)^{2}-3\sqrt{2}\sqrt{6}$$

Multiply 2 and 6 to get 12.

$$12\sqrt{3}+12-9\left(\sqrt{2}\right)^{2}-3\sqrt{2}\sqrt{6}$$

The square of \sqrt{2} is 2.

$$12\sqrt{3}+12-9\times 2-3\sqrt{2}\sqrt{6}$$

Multiply -9 and 2 to get -18.

$$12\sqrt{3}+12-18-3\sqrt{2}\sqrt{6}$$

Subtract 18 from 12 to get -6.

$$12\sqrt{3}-6-3\sqrt{2}\sqrt{6}$$

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

$$12\sqrt{3}-6-3\sqrt{2}\sqrt{2}\sqrt{3}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$12\sqrt{3}-6-3\times 2\sqrt{3}$$

Multiply -3 and 2 to get -6.

$$12\sqrt{3}-6-6\sqrt{3}$$

Combine 12\sqrt{3} and -6\sqrt{3} to get 6\sqrt{3}.

$$6\sqrt{3}-6$$

Factor

$$6 {(\sqrt{3} - 1)} = 4.392304845$$