Solve (2sqrt{3}-2)*(sqrt{3}+1)-3divsqrt{27}-(sqrt{3}-1)^2

$$( 2 \sqrt { 3 } - 2 ) \times ( \sqrt { 3 } + 1 ) - 3 \div \sqrt { 27 } - ( \sqrt { 3 } - 1 ) ^ { 2 }$$

Evaluate

$$\frac{5\sqrt{3}}{3}\approx 2.886751346$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 in both numerator and denominator.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

$$\left(2\sqrt{3}-2\right)\left(\sqrt{3}+1\right)-\frac{\sqrt{3}}{3}-\left(4-2\sqrt{3}\right)$$

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

$$\left(2\sqrt{3}-2\right)\left(\sqrt{3}+1\right)-\frac{\sqrt{3}}{3}-4+2\sqrt{3}$$

Combine -\frac{\sqrt{3}}{3} and 2\sqrt{3} to get \frac{5}{3}\sqrt{3}.

$$\left(2\sqrt{3}-2\right)\left(\sqrt{3}+1\right)+\frac{5}{3}\sqrt{3}-4$$

Use the distributive property to multiply 2\sqrt{3}-2 by \sqrt{3}+1 and combine like terms.

$$2\left(\sqrt{3}\right)^{2}-2+\frac{5}{3}\sqrt{3}-4$$

The square of \sqrt{3} is 3.

$$2\times 3-2+\frac{5}{3}\sqrt{3}-4$$

Multiply 2 and 3 to get 6.

$$6-2+\frac{5}{3}\sqrt{3}-4$$

Subtract 2 from 6 to get 4.

$$4+\frac{5}{3}\sqrt{3}-4$$

Subtract 4 from 4 to get 0.

$$\frac{5}{3}\sqrt{3}$$