By what number should
Question:

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

Solution:

Let the number be $x$.

$\therefore\left(\frac{-2}{3}\right)^{-3} \div x=\left(\frac{4}{27}\right)^{-2}$

$\Rightarrow\left(\frac{3}{-2}\right)^{3} \div x=\left(\frac{27}{4}\right)^{2}$

$\Rightarrow\left(\frac{-3}{2}\right)^{3} \div x=\left(\frac{27}{4}\right)^{2}$

$\Rightarrow\left(\frac{-3}{2}\right)^{3} \times \frac{1}{x}=\left(\frac{27}{4}\right)^{2}$

$\Rightarrow \frac{-3^{3}}{2^{3}} \times \frac{1}{x}=\frac{27^{2}}{4^{2}}$

$\Rightarrow \frac{-27}{8} \times \frac{1}{x}=\frac{27^{2}}{4^{2}}=\frac{27 \times 27}{4 \times 4}=\frac{27 \times 27}{4 \times 2 \times 2}=\frac{27 \times 27}{8 \times 2}$

$\therefore \frac{1}{x}=\left(\frac{27 \times 27}{8 \times 2}\right) /\left(\frac{-27}{8}\right)$

$\Rightarrow x=\frac{\left(\frac{-27}{8}\right)}{\left(\frac{27 \times 27}{8 \times 2}\right)}=\left(\frac{-27}{8}\right) \times\left(\frac{8 \times 2}{27 \times 27}\right)=\frac{-2}{27}$

 

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