Question

Question 98(ii)

Simplify:
${\left({\left(\frac{-2}{3}\right)}^{-2}\right)}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}$

Answer 1

Using the laws of exponents,
$(a^m{)}^n=(a{)}^{m\times n},=a^m\times a^n=a^{m+n}and{a}^m÷a^n=a^{m-n}$

$\therefore {\left({\left(\frac{-2}{3}\right)}^{-2}\right)}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}$

$={\left(\frac{2}{3}\right)}^{-2\times 3}\times (3{)}^4\times \frac{1}{3}\times \frac{1}{6}$

$={\left(\frac{-2}{3}\right)}^{-6}\times 3^4\times \frac{1}{3}\times \frac{1}{2\times 3}$

$={\left(\frac{3}{-2}\right)}^6\times 3^4\times \frac{1}{3\times 2\times 3}=\frac{(3{)}^6}{(-2{)}^6}\times 3^4\times \frac{1}{2^1\times 3^2}$

​​​​​​​=$\frac{(3{)}^6\times (3{)}^4}{(-1{)}^6(2{)}^6\times (2{)}^1\times (3{)}^2}$

$=\frac{(3{)}^{6+4}}{(2{)}^{6+1}\times 3^2}=\frac{(3{)}^{10}}{2^7\times 3^2}$

$=\frac{3^{10-2}}{2^7}=\frac{3^8}{2^7}$