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

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

Evaluate

$$2\sqrt{3}\approx 3.464101615$$

Solution Steps

Use the distributive property to multiply 1+\sqrt{2}+\sqrt{3} by 2+\sqrt{2}-\sqrt{6} and combine like terms.

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

The square of \sqrt{2} is 2.

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

Add 2 and 2 to get 4.

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

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}.

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

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

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

Combine -2\sqrt{3} and 2\sqrt{3} to get 0.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

Combine -\sqrt{6} and \sqrt{6} to get 0.

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

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}.

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

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

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

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

To find the opposite of 4-2\sqrt{3}, find the opposite of each term.

$$4-4+2\sqrt{3}$$

Subtract 4 from 4 to get 0.

$$2\sqrt{3}$$