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

13 \(\frac{1}{3}\) 13 \(\frac{2}{3}\) 14 \(\frac{2}{3}\) 17 \(\frac{1}{3}\)

14 \(\frac{2}{3}\)

a : b = 2 : 3 a = 2, b = 3 Now, ⇒ \(\frac{a^{3}\;+\;2ab^{2}}{b^{3}\;-\;2a^{2}b}\) = \(\frac{2^{3}\;+\;2\;×\;2\;×\;3^{2}}{3^{3}\;-\;2\;×\;2^{2}\;×\;3}\) = \(\frac{8\;+\;36}{27\;-\;24}\) = \(\frac{44}{3}\) = 14\(\frac{2}{3}\)