Question

By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient is ${\left(\frac{4}{9}\right)}^{-2}$?

Answer 1

Let the number be $x$.

$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{9}\right)}^{-2}$

$$\begin{gathered} \Rightarrow {\left(\frac{3}{-2}\right)}^3÷x={\left(\frac{9}{4}\right)}^2 \\ \Rightarrow \frac{{\left(\frac{3}{-2}\right)}^3}{x}={\left(\frac{9}{4}\right)}^2 \\ \Rightarrow \frac{{\displaystyle \frac{3^3}{-2^3}}}{x}=\frac{9^2}{4^2} \\ \Rightarrow x=\frac{\left({\displaystyle \frac{3^3}{-2^3}}\right)}{\left({\displaystyle \frac{9^2}{4^2}}\right)}=\frac{\left({\displaystyle \frac{3^3}{-2^3}}\right)}{\left({\displaystyle \frac{{\left(3^2\right)}^2}{{\left(2^2\right)}^2}}\right)}=\left({\displaystyle \frac{3^3}{-2^3}}\right)\times \left({\displaystyle \frac{{\left(2^2\right)}^2}{{\left(3^2\right)}^2}}\right)=\left({\displaystyle \frac{3^3}{-2^3}}\right)\times \left({\displaystyle \frac{2^4}{3^4}}\right)=\left({\displaystyle \frac{3^3}{-2^3}}\right)\times \left({\displaystyle \frac{2^3}{3^3}}\right)\times \left(\frac{2^1}{3^1}\right) \\ \Rightarrow \left({\displaystyle \frac{1}{-1}}\right)\times \left(\frac{2^1}{3^1}\right)=\frac{2}{-3}=\frac{2\times -1}{-3\times -1}=\frac{-2}{3} \end{gathered}$$