Question

Evaluate:
(i) ${\left\{{\left(\frac{-2}{3}\right)}^2\right\}}^{-2}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^2\right\}}^{-2}\right]}^{-1}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^2$

Answer 1

(i) ${\left\{{\left(\frac{-2}{3}\right)}^2\right\}}^{-2}={\left(\frac{-2}{3}\right)}^{2\times \left(-2\right)}={\left(\frac{-2}{3}\right)}^{-4}={\left(\frac{3}{-2}\right)}^4=\frac{3^4}{(-2{)}^4}=\frac{3^4}{2^4}=\frac{81}{16}$
(ii) ${\left[{\left\{{\left(\frac{-1}{3}\right)}^2\right\}}^{-2}\right]}^{-1}={\left[{\left(\frac{-1}{3}\right)}^{2\times \left(-2\right)}\right]}^{-1}={\left[{\left(\frac{-1}{3}\right)}^{-4}\right]}^{-1}={\left(\frac{-1}{3}\right)}^{-4\times -1}={\left(\frac{-1}{3}\right)}^4=\frac{-1^4}{3^4}=\frac{1^4}{3^4}=\frac{1}{81}$
(iii) ${\left\{{\left(\frac{3}{2}\right)}^{-2}\right\}}^2={\left(\frac{3}{2}\right)}^{-2\times 2}={\left(\frac{3}{2}\right)}^{-4}={\left(\frac{2}{3}\right)}^4=\frac{2^4}{3^4}=\frac{16}{81}$