Simplify:

Question:

Simplify:

(i) $\left(3^{2}+2^{2}\right) \times\left(\frac{1}{2}\right)^{3}$

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

(iii) $\left[\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right] \div\left(\frac{1}{4}\right)^{-3}$

(iv) $\left(2^{2}+3^{2}-4^{2}\right) \div\left(\frac{3}{2}\right)^{2}$

Solution:

(i) $\left(3^{2}+2^{2}\right) \times\left(\frac{1}{2}\right)^{3}=(9+4) \times \frac{1}{8}=\frac{13}{8}$

(ii) $\left(3^{2}-2^{2}\right) \times\left(\frac{2}{3}\right)^{-3}=(9-4) \times \frac{1}{(2 / 3)^{3}}$      ---> (a−1=1/(an))

$=5 \times \frac{1}{8 / 27}$   ---> ((a/b)n = (an)/(bn))

$=5 \times \frac{27}{8}$

$=\frac{135}{8}$

(iii) $\left(\left(\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right) \div\left(\frac{1}{4}\right)^{-3}=\left(3^{3}-2^{3}\right) \div 4^{3} \quad \ldots\left(a^{-n}=1 /\left(a^{n}\right)\right)\right.$

$=(27-8) \div 64$

$=19 \times \frac{1}{64}$

$=\frac{19}{64}$

(iv) $\left(2^{2}+3^{2}-4^{2}\right) \div\left(\frac{3}{2}\right)^{2}=(4+9-16) \times \frac{9}{4} \quad \cdots\left((a / b)^{n}=\left(a^{n}\right) /\left(b^{n}\right)\right)$

$=-3 \times \frac{9}{4}$

$=\frac{-27}{4}$