Question

Q2. What is the value of the expression?

Cube root under (-64) / square root under (-4)

Answer 1

$$\begin{gathered} \frac{{\left(-64\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(-4\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{{\left(-4\times -4\times -4\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(-4\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{{\left[{\left(-4\right)}^3\right]}^{{\displaystyle \frac{1}{3}}}}{{\left(-4\right)}^{\frac{1}{2}}} \\ =\frac{-4^{3\times {\displaystyle \frac{1}{3}}}}{{\left(-4\right)}^{\frac{1}{2}}}=\frac{-4^1}{{\left(-4\right)}^{\frac{1}{2}}}=-4^{1-\frac{1}{2}}\left[\mathrm{Since}\mathrm{base}\mathrm{are}\mathrm{same}\mathrm{so}\mathrm{we}\mathrm{can}\mathrm{subtract}\mathrm{powers}\right] \\ =-4^{\frac{1}{2}} \\ =\sqrt{-4} \end{gathered}$$