Question

Simplify, giving solution with positive index​ ; (xx) $(\frac{1}{4a{b}^{2}c}{)}^2÷(\frac{3}{2a^{2}b{c}^2}{)}^4$

Answer 1

Given, $(\frac{1}{4a{b}^{2}c}{)}^2÷(\frac{3}{2a^{2}b{c}^2}{)}^4$

It can be written as:
$(\frac{1}{4a{b}^{2}c}{)}^2÷(\frac{3}{2a^{2}b{c}^2}{)}^4=(\frac{1}{4a{b}^{2}c}{)}^2\times (\frac{2a^{2}b{c}^2}{3}{)}^4$

On further calculation , we have :
$(\frac{1}{4a{b}^{2}c}{)}^2÷(\frac{3}{2a^{2}b{c}^2}{)}^4$ $=\frac{1}{4^2{a}^2{b}^{2\times 2}{c}^2}\times \frac{2^4{a}^{2\times 4}{b}^4{c}^{2\times 4}}{3^4}$

[$\because$$(\frac{a}{b}{)}^m=\frac{a^m}{b^m},(ab{)}^m=a^m{b}^m,(a^m{)}^n=a^{mn}$]

$=\frac{2^4}{4^2\left(3^4\right)}\times a^{8-2}{b}^{4-4}{c}^{8-2}$

$=\frac{1}{3\times 3\times 3\times 3}{a}^6{b}^0{c}^6$

$=\frac{1}{81}{a}^6{c}^6$

Hence, $(\frac{1}{4ab{b}^{2}c}{)}^2÷(\frac{3}{2a^{2}b{c}^2}{)}^4=\frac{1}{81}{a}^6{c}^6$