Question

By what number should ${\left(\frac{-3}{2}\right)}^{-3}$ be divided so that the quotient may be ${\left(\frac{-8}{27}\right)}^{-2}$

Answer 1

Let x be the number which divides $(\frac{-3}{2}{)}^{-3}$ and given the quotient $(\frac{-8}{27}{)}^{-2}$

$\therefore (\frac{-3}{2}{)}^{-3}=x(\frac{-8}{27}{)}^{-2}$

$(\frac{-3}{2}{)}^{-3}=x(\frac{-2^3}{3^3}{)}^{-2}$

$(\frac{-3}{2}{)}^{-3}=x(\frac{-2}{3}{)}^{3.(-2)}$

$\Rightarrow \frac{1}{(\frac{-3}{2}{)}^3}=x.(\frac{-2}{3}{)}^{-6}$

$\Rightarrow \frac{1}{-\frac{3^3}{2^3}}=x.\frac{1}{(\frac{-2}{3}{)}^6}$

$\Rightarrow \frac{-2^3}{3^3}=x.\frac{1}{\frac{2^6}{3^6}}$

$\Rightarrow \frac{-2^3}{3^3}.\frac{2^6}{3^6}=x$

$\Rightarrow x=\frac{-2^{3+6}}{3^{3+6}}$ $[\because a^m.a^n=a^{m+n}]$

$\Rightarrow x=\frac{-2^9}{3^9}$

$\Rightarrow x=(-1{)}^9\frac{2^9}{3^9}$

$\Rightarrow x=(\frac{-2}{3}{)}^9$

$\therefore$ The required number = $(\frac{-2}{3}{)}^9$