Solve (sqrt{6}-2sqrt{3})^2-(sqrt{2}+2sqrt{5})(2sqrt{5}-sqrt{2})

$$( \sqrt { 6 } - 2 \sqrt { 3 } ) ^ { 2 } - ( \sqrt { 2 } + 2 \sqrt { 5 } ) ( 2 \sqrt { 5 } - \sqrt { 2 } )$$

Evaluate

$$-12\sqrt{2}\approx -16.970562748$$

Solution Steps

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

$$\left(\sqrt{6}\right)^{2}-4\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

The square of \sqrt{6} is 6.

$$6-4\sqrt{6}\sqrt{3}+4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

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}.

$$6-4\sqrt{3}\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

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

$$6-4\times 3\sqrt{2}+4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

Multiply -4 and 3 to get -12.

$$6-12\sqrt{2}+4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

The square of \sqrt{3} is 3.

$$6-12\sqrt{2}+4\times 3-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

Multiply 4 and 3 to get 12.

$$6-12\sqrt{2}+12-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

Add 6 and 12 to get 18.

$$18-12\sqrt{2}-\left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right)$$

Consider \left(\sqrt{2}+2\sqrt{5}\right)\left(2\sqrt{5}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$18-12\sqrt{2}-\left(\left(2\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}\right)$$

Expand \left(2\sqrt{5}\right)^{2}.

$$18-12\sqrt{2}-\left(2^{2}\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$18-12\sqrt{2}-\left(4\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$18-12\sqrt{2}-\left(4\times 5-\left(\sqrt{2}\right)^{2}\right)$$

Multiply 4 and 5 to get 20.

$$18-12\sqrt{2}-\left(20-\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$18-12\sqrt{2}-\left(20-2\right)$$

Subtract 2 from 20 to get 18.

$$18-12\sqrt{2}-18$$

Subtract 18 from 18 to get 0.

$$-12\sqrt{2}$$