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

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

Evaluate

$$-4\sqrt{3}-6\approx -12.92820323$$

Solution Steps

Consider \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

The square of \sqrt{5} is 5.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 5 to get 2.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

$$2-\left(8+4\sqrt{3}\right)$$

To find the opposite of 8+4\sqrt{3}, find the opposite of each term.

$$2-8-4\sqrt{3}$$

Subtract 8 from 2 to get -6.

$$-6-4\sqrt{3}$$