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

Explanation

$\left(-\frac{2}{3}\right)^{-3} = \left(-\frac{3}{2}\right)^3$ and $\left(\frac{4}{9}\right)^{-2} = \left(\frac{9}{4}\right)^2$; Now let the number asked in the question be x. Therefore, according to the question, $$\frac{\left(-\frac{3}{2}\right)^3}{x} = \left(\frac{9}{4}\right)^2$$ Therefore, $\left(-\frac{3}{2}\right)^3 = \left(\frac{9}{4}\right)^2 \times x$ Thus,, $x = \frac{-\frac{3}{2} \times -\frac{3}{2} \times -\frac{3}{2}}{\frac{9}{4} \times \frac{9}{4}} = -\frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{4}{9} \times \frac{4}{9}$ Thus, $x = -\frac{2}{3}$ Therefore, on dividing $\left(-\frac{2}{3}\right)^{-3}$ by $\left(-\frac{2}{3}\right)$, the quotient will be $\left(\frac{4}{9}\right)^{-2}$.