Solve (sqrt{3}-sqrt{6})^2-(1+2sqrt{2})^2+(1-sqrt{18})(1+sqrt{8})+sqrt{800}+11

$$( \sqrt { 3 } - \sqrt { 6 } ) ^ { 2 } - ( 1 + 2 \sqrt { 2 } ) ^ { 2 } + ( 1 - \sqrt { 18 } ) ( 1 + \sqrt { 8 } ) + \sqrt { 800 } + 11$$

Evaluate

$$9\sqrt{2}\approx 12.727922061$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{6}\right)^{2}.

$$\left(\sqrt{3}\right)^{2}-2\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

The square of \sqrt{3} is 3.

$$3-2\sqrt{3}\sqrt{6}+\left(\sqrt{6}\right)^{2}-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

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}.

$$3-2\sqrt{3}\sqrt{3}\sqrt{2}+\left(\sqrt{6}\right)^{2}-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

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

$$3-2\times 3\sqrt{2}+\left(\sqrt{6}\right)^{2}-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Multiply -2 and 3 to get -6.

$$3-6\sqrt{2}+\left(\sqrt{6}\right)^{2}-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

The square of \sqrt{6} is 6.

$$3-6\sqrt{2}+6-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Add 3 and 6 to get 9.

$$9-6\sqrt{2}-\left(1+2\sqrt{2}\right)^{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

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

$$9-6\sqrt{2}-\left(1+4\sqrt{2}+4\left(\sqrt{2}\right)^{2}\right)+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

The square of \sqrt{2} is 2.

$$9-6\sqrt{2}-\left(1+4\sqrt{2}+4\times 2\right)+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Multiply 4 and 2 to get 8.

$$9-6\sqrt{2}-\left(1+4\sqrt{2}+8\right)+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Add 1 and 8 to get 9.

$$9-6\sqrt{2}-\left(9+4\sqrt{2}\right)+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

To find the opposite of 9+4\sqrt{2}, find the opposite of each term.

$$9-6\sqrt{2}-9-4\sqrt{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Subtract 9 from 9 to get 0.

$$-6\sqrt{2}-4\sqrt{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Combine -6\sqrt{2} and -4\sqrt{2} to get -10\sqrt{2}.

$$-10\sqrt{2}+\left(1-\sqrt{18}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$-10\sqrt{2}+\left(1-3\sqrt{2}\right)\left(1+\sqrt{8}\right)+\sqrt{800}+11$$

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}.

$$-10\sqrt{2}+\left(1-3\sqrt{2}\right)\left(1+2\sqrt{2}\right)+\sqrt{800}+11$$

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

$$-10\sqrt{2}+1-\sqrt{2}-6\left(\sqrt{2}\right)^{2}+\sqrt{800}+11$$

The square of \sqrt{2} is 2.

$$-10\sqrt{2}+1-\sqrt{2}-6\times 2+\sqrt{800}+11$$

Multiply -6 and 2 to get -12.

$$-10\sqrt{2}+1-\sqrt{2}-12+\sqrt{800}+11$$

Subtract 12 from 1 to get -11.

$$-10\sqrt{2}-11-\sqrt{2}+\sqrt{800}+11$$

Combine -10\sqrt{2} and -\sqrt{2} to get -11\sqrt{2}.

$$-11\sqrt{2}-11+\sqrt{800}+11$$

Factor 800=20^{2}\times 2. Rewrite the square root of the product \sqrt{20^{2}\times 2} as the product of square roots \sqrt{20^{2}}\sqrt{2}. Take the square root of 20^{2}.

$$-11\sqrt{2}-11+20\sqrt{2}+11$$

Combine -11\sqrt{2} and 20\sqrt{2} to get 9\sqrt{2}.

$$9\sqrt{2}-11+11$$

Add -11 and 11 to get 0.

$$9\sqrt{2}$$