Solve (3+sqrt{8})(3-sqrt{8})[(3+sqrt{8}-(3-sqrt{8})]

$$( 3 + \sqrt { 8 } ) ( 3 - \sqrt { 8 } ) [ ( 3 + \sqrt { 8 } - ( 3 - \sqrt { 8 } ) ]$$

Evaluate

$$4\sqrt{2}\approx 5.656854249$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

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

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

Subtract 3 from 3 to get 0.

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

The opposite of -2\sqrt{2} is 2\sqrt{2}.

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

Combine 2\sqrt{2} and 2\sqrt{2} to get 4\sqrt{2}.

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

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

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

Combine -6\sqrt{2} and 6\sqrt{2} to get 0.

$$\left(9-4\left(\sqrt{2}\right)^{2}\right)\times 4\sqrt{2}$$

The square of \sqrt{2} is 2.

$$\left(9-4\times 2\right)\times 4\sqrt{2}$$

Multiply -4 and 2 to get -8.

$$\left(9-8\right)\times 4\sqrt{2}$$

Subtract 8 from 9 to get 1.

$$1\times 4\sqrt{2}$$

Multiply 1 and 4 to get 4.

$$4\sqrt{2}$$