Question

can any one solve this problem ​

Answer 1

Answer:

$\frac{19}{64}$

Step-by-step explanation:

$\frac{(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}}{(\frac{1}{4})^{-3}}$

Exponent rule (a/b)^c = (b/a)^c

$\frac{(\frac{3}{1})^{3}-(\frac{2}{1})^{3}}{(\frac{4}{1})^{3}}$

$\frac{3^3-2^2}{4^3}$

$\frac{27-8}{64}$

$\frac{19}{64}$