Question

If simplified form of $\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\}÷{\left(\frac{1}{4}\right)}^{-3}$ is $\frac{a}{b}$, where $a$ and $b$ are co-primes, then $b-3a$ is equal to:

Answer 1

$\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\}÷{\left(\frac{1}{4}\right)}^{-3}$
$=\{{\left(3\right)}^3-{\left(2\right)}^3\}÷{\left(4\right)}^3$
$=\{27-8\}÷\left(64\right)$
$=\left\{19\right\}÷\left(64\right)$
$=\frac{19}{64}$

Hence, $a=19$ and $b=64$

$\therefore b-3a=64-3\times 19=64-57=7$