Solve sqrt[3]{1-7/8}-(1/2-2)^3+(-2/3)^-3

$$\sqrt[ 3 ] { 1 - \frac { 7 } { 8 } } - ( \frac { 1 } { 2 } - 2 ) ^ { 3 } + ( - \frac { 2 } { 3 } ) ^ { - 3 }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Subtract \frac{7}{8} from 1 to get \frac{1}{8}.

$$\sqrt[3]{\frac{1}{8}}-\left(\frac{1}{2}-2\right)^{3}+\left(-\frac{2}{3}\right)^{-3}$$

Calculate \sqrt[3]{\frac{1}{8}} and get \frac{1}{2}.

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

Subtract 2 from \frac{1}{2} to get -\frac{3}{2}.

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

Calculate -\frac{3}{2} to the power of 3 and get -\frac{27}{8}.

$$\frac{1}{2}-\left(-\frac{27}{8}\right)+\left(-\frac{2}{3}\right)^{-3}$$

The opposite of -\frac{27}{8} is \frac{27}{8}.

$$\frac{1}{2}+\frac{27}{8}+\left(-\frac{2}{3}\right)^{-3}$$

Add \frac{1}{2} and \frac{27}{8} to get \frac{31}{8}.

$$\frac{31}{8}+\left(-\frac{2}{3}\right)^{-3}$$

Calculate -\frac{2}{3} to the power of -3 and get -\frac{27}{8}.

$$\frac{31}{8}-\frac{27}{8}$$

Subtract \frac{27}{8} from \frac{31}{8} to get \frac{1}{2}.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$