Question

Mark (✓) against the correct answer
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=?$

(a) $\frac{19}{64}$
(b) $\frac{64}{19}$
(c) $\frac{27}{16}$
(d) none of these

Answer 1

(a) $\frac{19}{64}$
We have:
$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}$ = $\left\{{\left(\frac{3}{1}\right)}^3-{\left(\frac{2}{1}\right)}^3\right\}÷{\left(\frac{4}{1}\right)}^3$ $\left[s\mathrm{ince}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-1}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^1\right]$
= $\left\{\left(3^3\right)-{\left(2\right)}^3\right\}÷{\left(4\right)}^3$
= $\left(27-8\right)÷64$
= $19÷64$
= $19\times \frac{1}{64}=\frac{19}{64}$