Solve 42(1+sqrt{2}-sqrt{3})(1-sqrt{2}+sqrt{3})

$$42 ( 1 + \sqrt { 2 } - \sqrt { 3 } ) ( 1 - \sqrt { 2 } + \sqrt { 3 } )$$

Evaluate

$$84\left(\sqrt{6}-2\right)\approx 37.757138394$$

Solution Steps

Use the distributive property to multiply 42 by 1+\sqrt{2}-\sqrt{3}.

$$\left(42+42\sqrt{2}-42\sqrt{3}\right)\left(1-\sqrt{2}+\sqrt{3}\right)$$

Apply the distributive property by multiplying each term of 42+42\sqrt{2}-42\sqrt{3} by each term of 1-\sqrt{2}+\sqrt{3}.

$$42-42\sqrt{2}+42\sqrt{3}+42\sqrt{2}-42\left(\sqrt{2}\right)^{2}+42\sqrt{2}\sqrt{3}-42\sqrt{3}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

Combine -42\sqrt{2} and 42\sqrt{2} to get 0.

$$42+42\sqrt{3}-42\left(\sqrt{2}\right)^{2}+42\sqrt{2}\sqrt{3}-42\sqrt{3}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$42+42\sqrt{3}-42\times 2+42\sqrt{2}\sqrt{3}-42\sqrt{3}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

Multiply -42 and 2 to get -84.

$$42+42\sqrt{3}-84+42\sqrt{2}\sqrt{3}-42\sqrt{3}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

Subtract 84 from 42 to get -42.

$$-42+42\sqrt{3}+42\sqrt{2}\sqrt{3}-42\sqrt{3}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

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

$$-42+42\sqrt{3}+42\sqrt{6}-42\sqrt{3}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

Combine 42\sqrt{3} and -42\sqrt{3} to get 0.

$$-42+42\sqrt{6}+42\sqrt{3}\sqrt{2}-42\left(\sqrt{3}\right)^{2}$$

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

$$-42+42\sqrt{6}+42\sqrt{6}-42\left(\sqrt{3}\right)^{2}$$

Combine 42\sqrt{6} and 42\sqrt{6} to get 84\sqrt{6}.

$$-42+84\sqrt{6}-42\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$-42+84\sqrt{6}-42\times 3$$

Multiply -42 and 3 to get -126.

$$-42+84\sqrt{6}-126$$

Subtract 126 from -42 to get -168.

$$-168+84\sqrt{6}$$