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

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

Evaluate

$$4\sqrt{3}\approx 6.92820323$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

$$\sqrt{3}+1+\sqrt{27}-\left(\sqrt{3}-1\right)^{0}$$

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}.

$$\sqrt{3}+1+3\sqrt{3}-\left(\sqrt{3}-1\right)^{0}$$

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

$$4\sqrt{3}+1-\left(\sqrt{3}-1\right)^{0}$$

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

$$4\sqrt{3}+1-1$$

Subtract 1 from 1 to get 0.

$$4\sqrt{3}$$