Question

By what number should ${\left(\frac{-2}{3}\right)}^{-3}$ be divided so that the quotient may be ${\left(\frac{4}{27}\right)}^{-2}$?

Answer 1

Let the number be $x$.

$\therefore$ ${\left(\frac{-2}{3}\right)}^{-3}÷x={\left(\frac{4}{27}\right)}^{-2}$
$$\begin{gathered} \Rightarrow {\left(\frac{3}{-2}\right)}^3÷x={\left(\frac{27}{4}\right)}^2 \\ \Rightarrow {\left(\frac{-3}{2}\right)}^3÷x={\left(\frac{27}{4}\right)}^2 \\ \Rightarrow {\left(\frac{-3}{2}\right)}^3\times \frac{1}{x}={\left(\frac{27}{4}\right)}^2 \\ \Rightarrow \frac{-3^3}{2^3}\times \frac{1}{x}=\frac{{27}^2}{4^2} \\ \Rightarrow \frac{-27}{8}\times \frac{1}{x}=\frac{{27}^2}{4^2}=\frac{27\times 27}{4\times 4}=\frac{27\times 27}{4\times 2\times 2}=\frac{27\times 27}{8\times 2} \\ \therefore \frac{1}{x}=\raisebox{1ex}{$\left({\displaystyle \frac{27\times 27}{8\times 2}}\right)$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle \frac{-27}{8}}\right)$}\right. \\ \Rightarrow x=\frac{\left({\displaystyle \frac{-27}{8}}\right)}{\left({\displaystyle \frac{27\times 27}{8\times 2}}\right)}=\left({\displaystyle \frac{-27}{8}}\right)\times \left(\frac{8\times 2}{27\times 27}\right)=\frac{-2}{27} \end{gathered}$$