Question

Simplify, giving Solution with positive index

${\displaystyle {\left(\frac{1}{{\text{4ab}}^2\text{c}}\right)}^2÷{\left(\frac{3}{2{\text{a}}^2{\text{bc}}^2}\right)}^4}$

Answer 1

${\displaystyle {\left(\frac{1}{{\text{4ab}}^2\text{c}}\right)}^2÷{\left(\frac{3}{2{\text{a}}^2{\text{bc}}^2}\right)}^4}$

${\displaystyle ={\left(\frac{1}{{\text{4ab}}^2}\text{c}\right)}^2\times {\left(\frac{2{\text{a}}^2{\text{bc}}^2}{3}\right)}^4}$

${\displaystyle =\frac{1^2}{4^2{\text{a}}^2{\text{b}}^{2\times 2}.{\text{c}}^2}\times \frac{2^4{\text{a}}^{2\times 4}.{\text{b}}^4.{\text{c}}^{2\times 4}}{3^4}}$

${\displaystyle =\frac{1^2}{3^4}\times {\text{a}}^{8-2}{\text{b}}^{4-4}{\text{c}}^{8-2}}$   (∵ 2⁴ = 4²)

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

${\displaystyle =\frac{1}{81}{\text{a}}^6{\text{c}}^6}$   (∵ b⁰ = 1)