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

\(\frac{5}{4}\) \(\frac{28}{9}\) \(\frac{14}{9}\) \(\frac{13}{8}\)

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

a : b = 4 : 3 a = 4 , b = 3 Now, ⇒ \(\frac{a^{3}\;+\;3ab^{2}}{b^{3}\;-\;3a^{2}b}\) = \(\frac{1^{3}\;+\;3\;×\;1\;×\;3^{2}}{3^{3}\;-\;3\;×\;1^{2}\;×\;3}\) = \(\frac{1\;+\;27}{27\;-\;9}\) = \(\frac{28}{18}\) = \(\frac{14}{9}\)