Question

Simplify :$(\frac{1}{6^{-2}}-\frac{1}{3^{-3}})÷\frac{1}{5^{-2}}$

• A. $(\frac{3}{5}{)}^2$
• B. $(\frac{9}{25}{)}^2$
• C. $(\frac{5}{3}{)}^2$
• D. $(\frac{25}{9}{)}^2$

Answer 1

The correct option is A

$(\frac{3}{5}{)}^2$

We know that for a non-zero integer $a$ and any integer $m,$
$a^{-m}=\frac{1}{a^m}.$
$\therefore 6^{-2}=\frac{1}{6^2}$
$⟹\frac{1}{6^{-2}}=6^2=36$
Similarly, $\frac{1}{3^{-3}}=3^3=27$ and $\frac{1}{5^{-2}}=5^2=25.$
$⟹(\frac{1}{6^{-2}}-\frac{1}{3^{-3}})÷\frac{1}{5^{-2}}=(36-27)÷25$
$=\frac{9}{25}$
$=\frac{3^2}{5^2}$
$=\left(\frac{3}{5}{)}^2\right(\because \frac{a^m}{b^m}=(\frac{a}{b}{)}^m$