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

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

Evaluate

$$-4$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

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

Calculate the square root of 1 and get 1.

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

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Cancel out 3, the greatest common factor in 6 and 3.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$-4$$

Factor

$$-4$$