Solve sqrt{12}-sqrt{3}+sqrt{1/3}-sqrt[3]{27}

$$\sqrt{ 12 } - \sqrt{ 3 } + \sqrt{ \frac{ 1 }{ 3 } } - \sqrt[3]{ 27 }$$

Evaluate

$$\frac{4\sqrt{3}}{3}-3\approx -0.690598923$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$2\sqrt{3}-\sqrt{3}+\sqrt{\frac{1}{3}}-\sqrt[3]{27}$$

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

$$\sqrt{3}+\sqrt{\frac{1}{3}}-\sqrt[3]{27}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$\sqrt{3}+\frac{\sqrt{1}}{\sqrt{3}}-\sqrt[3]{27}$$

Calculate the square root of 1 and get 1.

$$\sqrt{3}+\frac{1}{\sqrt{3}}-\sqrt[3]{27}$$

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

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

The square of \sqrt{3} is 3.

$$\sqrt{3}+\frac{\sqrt{3}}{3}-\sqrt[3]{27}$$

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

$$\frac{4}{3}\sqrt{3}-\sqrt[3]{27}$$

Calculate \sqrt[3]{27} and get 3.

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