Evaluate:

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) $\left\{\left(\frac{3}{2}\right)^{-2}\right\}^{2}$

 

Solution:

(i) $\left\{\left(\frac{-2}{3}\right)^{2}\right\}^{-2}=\left(\frac{-2}{3}\right)^{2 \times(-2)}=\left(\frac{-2}{3}\right)^{-4}=\left(\frac{3}{-2}\right)^{4}=\frac{3^{4}}{(-2)^{4}}=\frac{3^{4}}{2^{4}}=\frac{81}{16}$

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

(iii) $\left\{\left(\frac{3}{2}\right)^{-2}\right\}^{2}=\left(\frac{3}{2}\right)^{-2 \times 2}=\left(\frac{3}{2}\right)^{-4}=\left(\frac{2}{3}\right)^{4}=\frac{2^{4}}{3^{4}}=\frac{16}{81}$

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