Solve (2)sqrt{48}div(2sqrt{3}-1/3sqrt{27})-(sqrt{3}-1)^2

$$( 2 ) \sqrt { 48 } \div ( 2 \sqrt { 3 } - \frac { 1 } { 3 } \sqrt { 27 } ) - ( \sqrt { 3 } - 1 ) ^ { 2 }$$

Evaluate

$$2\left(\sqrt{3}+2\right)\approx 7.464101615$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{2\times 4\sqrt{3}}{2\sqrt{3}-\frac{1}{3}\sqrt{27}}-\left(\sqrt{3}-1\right)^{2}$$

Multiply 2 and 4 to get 8.

$$\frac{8\sqrt{3}}{2\sqrt{3}-\frac{1}{3}\sqrt{27}}-\left(\sqrt{3}-1\right)^{2}$$

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}.

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

Multiply -\frac{1}{3} and 3 to get -1.

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

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

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

Rationalize the denominator of \frac{8\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply 8 and 3 to get 24.

$$\frac{24}{3}-\left(\sqrt{3}-1\right)^{2}$$

Divide 24 by 3 to get 8.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

$$8-\left(4-2\sqrt{3}\right)$$

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

$$8-4+2\sqrt{3}$$

Subtract 4 from 8 to get 4.

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