Solve (sqrt{6}-sqrt{2})^2[1+(2+sqrt{3})]^2

$$( \sqrt { 6 } - \sqrt { 2 } ) ^ { 2 } [ 1 + ( 2 + \sqrt { 3 } ) ] ^ { 2 }$$

Evaluate

$$24$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

$$\left(\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)\left(1+2+\sqrt{3}\right)^{2}$$

The square of \sqrt{6} is 6.

$$\left(6-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)\left(1+2+\sqrt{3}\right)^{2}$$

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}.

$$\left(6-2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)\left(1+2+\sqrt{3}\right)^{2}$$

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

$$\left(6-2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)\left(1+2+\sqrt{3}\right)^{2}$$

Multiply -2 and 2 to get -4.

$$\left(6-4\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)\left(1+2+\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\left(6-4\sqrt{3}+2\right)\left(1+2+\sqrt{3}\right)^{2}$$

Add 6 and 2 to get 8.

$$\left(8-4\sqrt{3}\right)\left(1+2+\sqrt{3}\right)^{2}$$

Add 1 and 2 to get 3.

$$\left(8-4\sqrt{3}\right)\left(3+\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{3}\right)^{2}.

$$\left(8-4\sqrt{3}\right)\left(9+6\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$\left(8-4\sqrt{3}\right)\left(9+6\sqrt{3}+3\right)$$

Add 9 and 3 to get 12.

$$\left(8-4\sqrt{3}\right)\left(12+6\sqrt{3}\right)$$

Use the distributive property to multiply 8-4\sqrt{3} by 12+6\sqrt{3} and combine like terms.

$$96-24\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$96-24\times 3$$

Multiply -24 and 3 to get -72.

$$96-72$$

Subtract 72 from 96 to get 24.

$$24$$

Factor

$$2^{3}\times 3$$