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

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

Evaluate

$$\frac{25\sqrt{3}-48}{11}\approx -0.427157256$$

Short Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}\left(1-\sqrt{3}\right)}{1+3\sqrt{3}} by multiplying numerator and denominator by 1-3\sqrt{3}.

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

Consider \left(1+3\sqrt{3}\right)\left(1-3\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 1 to the power of 2 and get 1.

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

Expand \left(3\sqrt{3}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{2\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-9\left(\sqrt{3}\right)^{2}}}{1-\frac{3\left(1-\sqrt{3}\right)^{2}}{\left(1+3\sqrt{3}\right)^{2}}}$$

The square of \sqrt{3} is 3.

$$\frac{\frac{2\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-9\times 3}}{1-\frac{3\left(1-\sqrt{3}\right)^{2}}{\left(1+3\sqrt{3}\right)^{2}}}$$

Multiply 9 and 3 to get 27.

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

Subtract 27 from 1 to get -26.

$$\frac{\frac{2\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{-26}}{1-\frac{3\left(1-\sqrt{3}\right)^{2}}{\left(1+3\sqrt{3}\right)^{2}}}$$

Divide 2\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right) by -26 to get -\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right).

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(1-\sqrt{3}\right)^{2}}{\left(1+3\sqrt{3}\right)^{2}}}$$

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

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(1-2\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)}{\left(1+3\sqrt{3}\right)^{2}}}$$

The square of \sqrt{3} is 3.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(1-2\sqrt{3}+3\right)}{\left(1+3\sqrt{3}\right)^{2}}}$$

Add 1 and 3 to get 4.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)}{\left(1+3\sqrt{3}\right)^{2}}}$$

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

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)}{1+6\sqrt{3}+9\left(\sqrt{3}\right)^{2}}}$$

The square of \sqrt{3} is 3.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)}{1+6\sqrt{3}+9\times 3}}$$

Multiply 9 and 3 to get 27.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)}{1+6\sqrt{3}+27}}$$

Add 1 and 27 to get 28.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)}{28+6\sqrt{3}}}$$

Rationalize the denominator of \frac{3\left(4-2\sqrt{3}\right)}{28+6\sqrt{3}} by multiplying numerator and denominator by 28-6\sqrt{3}.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{\left(28+6\sqrt{3}\right)\left(28-6\sqrt{3}\right)}}$$

Consider \left(28+6\sqrt{3}\right)\left(28-6\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{28^{2}-\left(6\sqrt{3}\right)^{2}}}$$

Calculate 28 to the power of 2 and get 784.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{784-\left(6\sqrt{3}\right)^{2}}}$$

Expand \left(6\sqrt{3}\right)^{2}.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{784-6^{2}\left(\sqrt{3}\right)^{2}}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{784-36\left(\sqrt{3}\right)^{2}}}$$

The square of \sqrt{3} is 3.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{784-36\times 3}}$$

Multiply 36 and 3 to get 108.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{784-108}}$$

Subtract 108 from 784 to get 676.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{1-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{676}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{676}{676}.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{\frac{676}{676}-\frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{676}}$$

Since \frac{676}{676} and \frac{3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{676} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{\frac{676-3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right)}{676}}$$

Do the multiplications in 676-3\left(4-2\sqrt{3}\right)\left(28-6\sqrt{3}\right).

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{\frac{676-336+72\sqrt{3}+168\sqrt{3}-108}{676}}$$

Do the calculations in 676-336+72\sqrt{3}+168\sqrt{3}-108.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{\frac{232+240\sqrt{3}}{676}}$$

Divide -\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right) by \frac{232+240\sqrt{3}}{676} by multiplying -\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right) by the reciprocal of \frac{232+240\sqrt{3}}{676}.

$$\frac{-\frac{1}{13}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\times 676}{232+240\sqrt{3}}$$

Multiply -\frac{1}{13} and 676 to get -52.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{232+240\sqrt{3}}$$

Rationalize the denominator of \frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)}{232+240\sqrt{3}} by multiplying numerator and denominator by 232-240\sqrt{3}.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{\left(232+240\sqrt{3}\right)\left(232-240\sqrt{3}\right)}$$

Consider \left(232+240\sqrt{3}\right)\left(232-240\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{232^{2}-\left(240\sqrt{3}\right)^{2}}$$

Calculate 232 to the power of 2 and get 53824.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{53824-\left(240\sqrt{3}\right)^{2}}$$

Expand \left(240\sqrt{3}\right)^{2}.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{53824-240^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 240 to the power of 2 and get 57600.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{53824-57600\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{53824-57600\times 3}$$

Multiply 57600 and 3 to get 172800.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{53824-172800}$$

Subtract 172800 from 53824 to get -118976.

$$\frac{-52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)}{-118976}$$

Divide -52\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right) by -118976 to get \frac{1}{2288}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right).

$$\frac{1}{2288}\sqrt{3}\left(1-\sqrt{3}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)$$

Use the distributive property to multiply \frac{1}{2288}\sqrt{3} by 1-\sqrt{3}.

$$\left(\frac{1}{2288}\sqrt{3}-\frac{1}{2288}\left(\sqrt{3}\right)^{2}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$\left(\frac{1}{2288}\sqrt{3}-\frac{1}{2288}\times 3\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)$$

Multiply -\frac{1}{2288} and 3 to get -\frac{3}{2288}.

$$\left(\frac{1}{2288}\sqrt{3}-\frac{3}{2288}\right)\left(1-3\sqrt{3}\right)\left(232-240\sqrt{3}\right)$$

Use the distributive property to multiply \frac{1}{2288}\sqrt{3}-\frac{3}{2288} by 1-3\sqrt{3} and combine like terms.

$$\left(\frac{5}{1144}\sqrt{3}-\frac{3}{2288}\left(\sqrt{3}\right)^{2}-\frac{3}{2288}\right)\left(232-240\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$\left(\frac{5}{1144}\sqrt{3}-\frac{3}{2288}\times 3-\frac{3}{2288}\right)\left(232-240\sqrt{3}\right)$$

Multiply -\frac{3}{2288} and 3 to get -\frac{9}{2288}.

$$\left(\frac{5}{1144}\sqrt{3}-\frac{9}{2288}-\frac{3}{2288}\right)\left(232-240\sqrt{3}\right)$$

Subtract \frac{3}{2288} from -\frac{9}{2288} to get -\frac{3}{572}.

$$\left(\frac{5}{1144}\sqrt{3}-\frac{3}{572}\right)\left(232-240\sqrt{3}\right)$$

Use the distributive property to multiply \frac{5}{1144}\sqrt{3}-\frac{3}{572} by 232-240\sqrt{3} and combine like terms.

$$\frac{25}{11}\sqrt{3}-\frac{150}{143}\left(\sqrt{3}\right)^{2}-\frac{174}{143}$$

The square of \sqrt{3} is 3.

$$\frac{25}{11}\sqrt{3}-\frac{150}{143}\times 3-\frac{174}{143}$$

Multiply -\frac{150}{143} and 3 to get -\frac{450}{143}.

$$\frac{25}{11}\sqrt{3}-\frac{450}{143}-\frac{174}{143}$$

Subtract \frac{174}{143} from -\frac{450}{143} to get -\frac{48}{11}.

$$\frac{25}{11}\sqrt{3}-\frac{48}{11}$$