Given that $\dfrac{a^3 + 3 a b^2}{b^3 + 3 a^2 b} = \dfrac{62}{63}$, find $a : b$ using componendo-dividendo.
Given: $\;$ $\dfrac{a^3 + 3 a b^2}{b^3 + 3 a^2 b} = \dfrac{62}{63}$
i.e. $\;$ $\dfrac{a^3 + 3 ab^2 + b^3 + 3 a^2 b}{a^3 + 3ab^2 - b^3 - 3a^2 b} = \dfrac{62 + 63}{62 - 63}$ $\;\;\;$ [by componendo-dividendo]
i.e. $\;$ $\dfrac{\left(a + b\right)^3}{\left(a - b\right)^3} = \dfrac{125}{-1}$
i.e. $\;$ $\dfrac{\left(a + b\right)^3}{\left(a - b\right)^3} = \dfrac{5^3}{-1^3}$
i.e. $\;$ $\dfrac{a + b}{a - b} = \dfrac{5}{-1}$
i.e. $\;$ $\dfrac{a + b + a - b}{a + b - a + b} = \dfrac{5 + \left(-1\right)}{5 - \left(-1\right)}$ $\;\;\;$ [by componendo-dividendo]
i.e. $\;$ $\dfrac{2a}{2b} = \dfrac{4}{6}$
i.e. $\;$ $\dfrac{a}{b} = \dfrac{2}{3}$