Question

Question 13

Cube of $\left(\frac{-1}{4}\right)is$
(a)$\frac{-1}{12}$
(b)$\frac{1}{16}$
(c)$\frac{-1}{64}$
(d)$\frac{1}{64}$

Answer 1

Cube of $\left(\frac{-1}{4}\right)\text{is}{\left(\frac{-1}{4}\right)}^3$
So, ${\left(\frac{-1}{4}\right)}^3=\left(\frac{-1}{4}\right)\times \left(\frac{-1}{4}\right)\times \left(\frac{-1}{4}\right)=\frac{(-1)\times (-1)\times (-1)}{4\times 4\times 4}=\frac{-1}{64}$