# By what number should

Question:

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

Solution:

Let the number be $x$.

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

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

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

$\Rightarrow \frac{3^{3}}{\frac{-2^{3}}{x}}=\frac{9^{2}}{4^{2}}$

$\Rightarrow x=\frac{\left(\frac{3^{3}}{-2^{3}}\right)}{\left(\frac{g^{2}}{4^{2}}\right)}=\frac{\left(\frac{3^{3}}{-2^{3}}\right)}{\left(\frac{\left(3^{2}\right)^{2}}{\left(2^{2}\right)^{2}}\right)}=\left(\frac{3^{3}}{-2^{3}}\right) \times\left(\frac{\left(2^{2}\right)^{2}}{\left(3^{2}\right)^{2}}\right)=\left(\frac{3^{3}}{-2^{3}}\right) \times\left(\frac{2^{4}}{3^{4}}\right)=\left(\frac{3^{3}}{-2^{3}}\right) \times\left(\frac{2^{3}}{3^{3}}\right) \times\left(\frac{2^{1}}{3^{1}}\right)$

$\Rightarrow\left(\frac{1}{-1}\right) \times\left(\frac{2^{1}}{3^{1}}\right)=\frac{2}{-3}=\frac{2 \times-1}{-3 \times-1}=\frac{-2}{3}$