Solve (6)(-sqrt{3})^0+(1/3)^-2+sqrt{27}-frac{2+sqrt{3}}{2-sqrt{3}}

$$( 6 ) ( - \sqrt { 3 } ) ^ { 0 } + ( \frac { 1 } { 3 } ) ^ { - 2 } + \sqrt { 27 } - \frac { 2 + \sqrt { 3 } } { 2 - \sqrt { 3 } }$$

Evaluate

$$8-\sqrt{3}\approx 6.267949192$$

Solution Steps

Calculate \frac{1}{3} to the power of -2 and get 9.

$$6\left(-\sqrt{3}\right)^{0}+9+\sqrt{27}-\frac{2+\sqrt{3}}{2-\sqrt{3}}$$

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}.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\frac{2+\sqrt{3}}{2-\sqrt{3}}$$

Rationalize the denominator of \frac{2+\sqrt{3}}{2-\sqrt{3}} by multiplying numerator and denominator by 2+\sqrt{3}.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}$$

Consider \left(2-\sqrt{3}\right)\left(2+\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}.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{2^{2}-\left(\sqrt{3}\right)^{2}}$$

Square 2. Square \sqrt{3}.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{4-3}$$

Subtract 3 from 4 to get 1.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)$$

Multiply 2+\sqrt{3} and 2+\sqrt{3} to get \left(2+\sqrt{3}\right)^{2}.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\left(2+\sqrt{3}\right)^{2}$$

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

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\left(4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\left(4+4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-\left(7+4\sqrt{3}\right)$$

To find the opposite of 7+4\sqrt{3}, find the opposite of each term.

$$6\left(-\sqrt{3}\right)^{0}+9+3\sqrt{3}-7-4\sqrt{3}$$

Subtract 7 from 9 to get 2.

$$6\left(-\sqrt{3}\right)^{0}+2+3\sqrt{3}-4\sqrt{3}$$

Combine 3\sqrt{3} and -4\sqrt{3} to get -\sqrt{3}.

$$6\left(-\sqrt{3}\right)^{0}+2-\sqrt{3}$$

Calculate -\sqrt{3} to the power of 0 and get 1.

$$6\times 1+2-\sqrt{3}$$

Multiply 6 and 1 to get 6.

$$6+2-\sqrt{3}$$

Add 6 and 2 to get 8.

$$8-\sqrt{3}$$