Solve sqrt[3]{-8}+sqrt{6}-(sqrt{2}+sqrt{2/3})

$$\sqrt[ 3 ]{ -8 } + \sqrt{ 6 } -( \sqrt{ 2 } + \sqrt{ \frac{ 2 }{ 3 } } )$$

Evaluate

$$\frac{2\sqrt{6}}{3}-\sqrt{2}-2\approx -1.781220401$$

Short Solution Steps

Calculate \sqrt[3]{-8} and get -2.

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

Rewrite the square root of the division \sqrt{\frac{2}{3}} as the division of square roots \frac{\sqrt{2}}{\sqrt{3}}.

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

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

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{3}{3}.

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

Since \frac{3\sqrt{2}}{3} and \frac{\sqrt{6}}{3} have the same denominator, add them by adding their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -2+\sqrt{6} times \frac{3}{3}.

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

Since \frac{3\left(-2+\sqrt{6}\right)}{3} and \frac{3\sqrt{2}+\sqrt{6}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(-2+\sqrt{6}\right)-\left(3\sqrt{2}+\sqrt{6}\right).

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

Do the calculations in -6+3\sqrt{6}-3\sqrt{2}-\sqrt{6}.

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