Solve sqrt[3]{1/3}+sqrt{12}-sqrt{48}

$$\sqrt[ 3 ] { \frac { 1 } { 3 } } + \sqrt { 12 } - \sqrt { 48 }$$

Evaluate

$$\frac{3^{\frac{2}{3}}-6\sqrt{3}}{3}\approx -2.770740341$$

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}.

$$\sqrt[3]{\frac{1}{3}}+2\sqrt{3}-\sqrt{48}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\sqrt[3]{\frac{1}{3}}+2\sqrt{3}-4\sqrt{3}$$

Combine 2\sqrt{3} and -4\sqrt{3} to get -2\sqrt{3}.

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