Solve frac{sqrt{3}(2sqrt{3}-6)}{3}-frac{(sqrt{3}-1)^2}{2}=

$$\frac { \sqrt { 3 } ( 2 \sqrt { 3 } - 6 ) } { 3 } - \frac { ( \sqrt { 3 } - 1 ) ^ { 2 } } { 2 } =$$

Evaluate

$$-\sqrt{3}\approx -1.732050808$$

Short Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Divide each term of 4-2\sqrt{3} by 2 to get 2-\sqrt{3}.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -2+\sqrt{3} times \frac{3}{3}.

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

Since \frac{\sqrt{3}\left(2\sqrt{3}-6\right)}{3} and \frac{3\left(-2+\sqrt{3}\right)}{3} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \sqrt{3}\left(2\sqrt{3}-6\right)+3\left(-2+\sqrt{3}\right).

$$\frac{6-6\sqrt{3}-6+3\sqrt{3}}{3}$$

Do the calculations in 6-6\sqrt{3}-6+3\sqrt{3}.

$$\frac{-3\sqrt{3}}{3}$$

Cancel out 3 and 3.

$$-\sqrt{3}$$