Solve -1^2020+left(sqrt{3}-sqrt{2}right)left(sqrt{2}+sqrt{3}right)-sqrt{8}divsqrt{1/6}

$$- { 1 }^{ 2020 } + \left( \sqrt{ 3 } - \sqrt{ 2 } \right) \left( \sqrt{ 2 } + \sqrt{ 3 } \right) - \sqrt{ 8 } \div \sqrt{ \frac{ 1 }{ 6 } }$$

Evaluate

$$-4\sqrt{3}\approx -6.92820323$$

Solution Steps

Calculate 1 to the power of 2020 and get 1.

$$-1+\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}\right)-\frac{\sqrt{8}}{\sqrt{\frac{1}{6}}}$$

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

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

The square of \sqrt{3} is 3.

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

The square of \sqrt{2} is 2.

$$-1+3-2-\frac{\sqrt{8}}{\sqrt{\frac{1}{6}}}$$

Subtract 2 from 3 to get 1.

$$-1+1-\frac{\sqrt{8}}{\sqrt{\frac{1}{6}}}$$

Add -1 and 1 to get 0.

$$-\frac{\sqrt{8}}{\sqrt{\frac{1}{6}}}$$

Rewrite the division of square roots \frac{\sqrt{8}}{\sqrt{\frac{1}{6}}} as the square root of the division \sqrt{\frac{8}{\frac{1}{6}}} and perform the division.

$$-\sqrt{8}$$

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}.

$$-2\sqrt{2}$$