Solve 3/sqrt{3}-{left(sqrt{3}right)}^2+left(pi+3right)^0-sqrt{27}+sqrt{3}-2

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

Evaluate

$$-\sqrt{3}-4\approx -5.732050808$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 and 3.

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

The square of \sqrt{3} is 3.

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

Calculate \pi +3 to the power of 0 and get 1.

$$\sqrt{3}-3+1-\sqrt{27}+\sqrt{3}-2$$

Add -3 and 1 to get -2.

$$\sqrt{3}-2-\sqrt{27}+\sqrt{3}-2$$

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}.

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

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

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

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

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

Subtract 2 from -2 to get -4.

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