If a : b = 1 : 3 , then find the value of (a^3+3ab^2)/(b^3- 3a^2b)

a : b = 4 : 3

a = 4 , b = 3

Now,

⇒ \(\frac{a^{3}\;+\;3ab^{2}}{b^{3}\;-\;3a^{2}b}\)

= \(\frac{1^{3}\;+\;3\;×\;1\;×\;3^{2}}{3^{3}\;-\;3\;×\;1^{2}\;×\;3}\)

= \(\frac{1\;+\;27}{27\;-\;9}\)

= \(\frac{28}{18}\)

= \(\frac{14}{9}\)