# Simplify:

Question:

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

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

(iii) $\left\{\left(\frac{1}{2}\right)^{-1} \times(-4)^{-1}\right\}^{-1}$

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

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

Solution:

(i) $\left(\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right) \div\left(\frac{1}{4}\right)^{-3}=\left(\frac{1}{(1 / 3)^{3}}-\frac{1}{(1 / 2)^{3}}\right) \div \frac{1}{(1 / 4)^{3}}$$\cdots\left(a^{-n}=1 /\left(a^{n}\right)\right) =\left(\frac{1}{(1 / 27)}-\frac{1}{(1 / 8)}\right) \div \frac{1}{(1 / 64)} =\left(\frac{27}{1}-\frac{8}{1}\right) \div 64 =(19) \times \frac{1}{64} =\frac{19}{64} (ii) \left(3^{2}-2^{2}\right) \times\left(\frac{2}{3}\right)^{-3}=(9-4) \times \frac{1}{(2 / 3)^{3}}$$\cdots\left(a^{-n}=1 /\left(a^{n}\right)\right)$

$=5 \times \frac{1}{8 / 27}$

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

$=\frac{135}{8}$

(iii) $\left(\left(\frac{1}{2}\right)^{-1} \times(-4)^{-1}\right)^{-1}=\left(\left(\frac{1}{1 / 2}\right) \times\left(\frac{1}{-4}\right)\right)^{-1}$ $\cdots\left(a^{-1}=1 / a\right)$

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

$=\left(\frac{1}{-2}\right)^{-1}$

$=\frac{1}{1 /(-2)} \quad \ldots\left(a^{-1}=1 / a\right)$

=-2

$(i v)\left(\left(\left(\frac{-1}{4}\right)^{2}\right)^{-2}\right)^{-1}=\left(\left(\frac{(-1)^{2}}{4^{2}}\right)^{-2}\right)^{-1}$$->\left((a / b)^{n}=a^{n} /\left(b^{n}\right)\right) =\left(\left(\frac{1}{16}\right)^{-2}\right)^{-1} \quad \stackrel{->}{\left(a^{-n}=1 /\left(a^{n}\right)\right)} =\left(\left(\frac{1}{(1 / 16)^{2}}\right)\right)^{-1} =\left(\frac{1}{(1 / 256)}\right)^{-1} =256^{-1} \quad \ldots\left(a^{-1}=1 / a\right) =\frac{1}{256} (v) \left(\left(\frac{2}{3}\right)^{2}\right)^{3} \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times 6^{-1}=\left(\frac{2^{2}}{3^{2}}\right)^{3} \times \frac{1}{(1 / 3)^{4}} \times \frac{1}{3} \times \frac{1}{6}$$\ldots\left((a / b)^{n}=a^{n} /\left(b^{n}\right)\right)$ and $\left(a^{-n}=1 /\left(a^{n}\right)\right)$

$=\left(\frac{4}{9}\right)^{3} \times \frac{1}{(1 / 81)} \times \frac{1}{3} \times \frac{1}{6}$

$=\frac{4^{3}}{9^{3}} \times 81 \times \frac{1}{18}$ \cdots\left((a / b)^{n}=a^{n} /\left(b^{n}\right)\right)

$=\frac{64}{729} \times 81 \times \frac{1}{18}$

$=\frac{64}{9} \times \frac{1}{18}$

$=64 \times \frac{1}{162}$

$=\frac{64}{162}$

$=\frac{32}{81}$

Leave a comment

Click here to get exam-ready with eSaral