Solve sqrt[3]{0125}-sqrt{31/16}+sqrt[3]{(1-7/8)^2}-|-1/2|

$$\sqrt[ 3 ] { 0125 } - \sqrt { 3 \frac { 1 } { 16 } } + \sqrt[ 3 ] { ( 1 - \frac { 7 } { 8 } ) ^ { 2 } } - | - \frac { 1 } { 2 } |$$

Evaluate

$$-2$$

Solution Steps

Multiply 0 and 125 to get 0.

$$\sqrt[3]{0}-\sqrt{\frac{3\times 16+1}{16}}+\sqrt[3]{\left(1-\frac{7}{8}\right)^{2}}-|-\frac{1}{2}|$$

Calculate \sqrt[3]{0} and get 0.

$$0-\sqrt{\frac{3\times 16+1}{16}}+\sqrt[3]{\left(1-\frac{7}{8}\right)^{2}}-|-\frac{1}{2}|$$

Multiply 3 and 16 to get 48.

$$0-\sqrt{\frac{48+1}{16}}+\sqrt[3]{\left(1-\frac{7}{8}\right)^{2}}-|-\frac{1}{2}|$$

Add 48 and 1 to get 49.

$$0-\sqrt{\frac{49}{16}}+\sqrt[3]{\left(1-\frac{7}{8}\right)^{2}}-|-\frac{1}{2}|$$

Rewrite the square root of the division \frac{49}{16} as the division of square roots \frac{\sqrt{49}}{\sqrt{16}}. Take the square root of both numerator and denominator.

$$0-\frac{7}{4}+\sqrt[3]{\left(1-\frac{7}{8}\right)^{2}}-|-\frac{1}{2}|$$

Subtract \frac{7}{4} from 0 to get -\frac{7}{4}.

$$-\frac{7}{4}+\sqrt[3]{\left(1-\frac{7}{8}\right)^{2}}-|-\frac{1}{2}|$$

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

$$-\frac{7}{4}+\sqrt[3]{\left(\frac{1}{8}\right)^{2}}-|-\frac{1}{2}|$$

Calculate \frac{1}{8} to the power of 2 and get \frac{1}{64}.

$$-\frac{7}{4}+\sqrt[3]{\frac{1}{64}}-|-\frac{1}{2}|$$

Calculate \sqrt[3]{\frac{1}{64}} and get \frac{1}{4}.

$$-\frac{7}{4}+\frac{1}{4}-|-\frac{1}{2}|$$

Add -\frac{7}{4} and \frac{1}{4} to get -\frac{3}{2}.

$$-\frac{3}{2}-|-\frac{1}{2}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{1}{2} is \frac{1}{2}.

$$-\frac{3}{2}-\frac{1}{2}$$

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

$$-2$$

Factor

$$-2$$