Solve (9-sqrt{3})^2-78/sqrt{3+9}

$$( 9 - \sqrt { 3 } ) ^ { 2 } - \frac { 78 } { \sqrt { 3 } + 9 }$$

Evaluate

$$75-17\sqrt{3}\approx 45.555136271$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-\sqrt{3}\right)^{2}.

$$81-18\sqrt{3}+\left(\sqrt{3}\right)^{2}-\frac{78}{\sqrt{3}+9}$$

The square of \sqrt{3} is 3.

$$81-18\sqrt{3}+3-\frac{78}{\sqrt{3}+9}$$

Add 81 and 3 to get 84.

$$84-18\sqrt{3}-\frac{78}{\sqrt{3}+9}$$

Rationalize the denominator of \frac{78}{\sqrt{3}+9} by multiplying numerator and denominator by \sqrt{3}-9.

$$84-18\sqrt{3}-\frac{78\left(\sqrt{3}-9\right)}{\left(\sqrt{3}+9\right)\left(\sqrt{3}-9\right)}$$

Consider \left(\sqrt{3}+9\right)\left(\sqrt{3}-9\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$84-18\sqrt{3}-\frac{78\left(\sqrt{3}-9\right)}{\left(\sqrt{3}\right)^{2}-9^{2}}$$

Square \sqrt{3}. Square 9.

$$84-18\sqrt{3}-\frac{78\left(\sqrt{3}-9\right)}{3-81}$$

Subtract 81 from 3 to get -78.

$$84-18\sqrt{3}-\frac{78\left(\sqrt{3}-9\right)}{-78}$$

Cancel out -78 and -78.

$$84-18\sqrt{3}-\left(-\left(\sqrt{3}-9\right)\right)$$

The opposite of -\left(\sqrt{3}-9\right) is \sqrt{3}-9.

$$84-18\sqrt{3}+\sqrt{3}-9$$

Combine -18\sqrt{3} and \sqrt{3} to get -17\sqrt{3}.

$$84-17\sqrt{3}-9$$

Subtract 9 from 84 to get 75.

$$75-17\sqrt{3}$$