Solve {l}{(1/3)^-1-21+sqrt{8}}{-(sqrt{3}-1)^2}

$$\left. \begin{array} { l } { ( \frac { 1 } { 3 } ) ^ { - 1 } - 21 + \sqrt { 8 } } \\ { - ( \sqrt { 3 } - 1 ) ^ { 2 } } \end{array} \right.$$

Evaluate

$$2\left(\sqrt{2}+\sqrt{3}-11\right)\approx -15.70747126$$

Solution Steps

Calculate \frac{1}{3} to the power of -1 and get 3.

$$3-21+\sqrt{8}-\left(\sqrt{3}-1\right)^{2}$$

Subtract 21 from 3 to get -18.

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

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}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

$$-18+2\sqrt{2}-\left(3-2\sqrt{3}+1\right)$$

Add 3 and 1 to get 4.

$$-18+2\sqrt{2}-\left(4-2\sqrt{3}\right)$$

To find the opposite of 4-2\sqrt{3}, find the opposite of each term.

$$-18+2\sqrt{2}-4+2\sqrt{3}$$

Subtract 4 from -18 to get -22.

$$-22+2\sqrt{2}+2\sqrt{3}$$