Question

Evaluate the expression ${\left(\sqrt[3]{-\frac{1}{8}}\right)}^3+\frac{3}{4}$ .

Answer 1

Given expression: ${\left(\sqrt[3]{-\frac{1}{8}}\right)}^3+\frac{3}{4}$.

We know that

$\sqrt[\mathrm{n}]{\mathrm{a}}={\left(\mathrm{a}\right)}^{\frac{1}{\mathrm{n}}}\&{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}\times \mathrm{n}}$

Now,

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

$$\begin{gathered} ={\left\{{\left(-\frac{1}{8}\right)}^{\frac{1}{3}}\right\}}^3+\frac{3}{4} \\ ={\left(-\frac{1}{8}\right)}^{\frac{1}{3}\times 3}+\frac{3}{4} \\ =-\frac{1}{8}+\frac{3}{4} \\ =\frac{-1+6}{8} \\ =\frac{5}{8} \end{gathered}$$

Hence, ${\left(\sqrt[3]{-\frac{1}{8}}\right)}^3+\frac{3}{4}=\frac{5}{8}$.