If a³ + 3a² + 9a = 1, then what is the value of $${a^3} + \frac{3}{a}?$$

Solution (By Examveda Team)

$$\eqalign{ & {a^3} + 3{a^2} + 9a = 1.......\left( {\text{i}} \right) \cr & {a^3} + \frac{3}{a} = ? \cr & {\text{Multiply by '}}a{\text{' and '3' in equation}}\left( {\text{i}} \right) \cr & \left( {{a^3} + 3{a^2} + 9a = 1} \right) \times a \cr & \left( {{a^3} + 3{a^2} + 9a = 1} \right) \times 3 \cr & {a^4} + 3{a^3} + 9{a^2} = 1 \times a \cr & \underline {3{a^3} + 9{a^2} + 27a = 1 \times 3} \to \left( {{\text{Subtracting}}} \right) \cr & {a^4} - 27a = a - 3 \cr & {a^4} + 3 = 28a \cr & {\text{On dividing by }}a \cr & {a^3} + \frac{3}{a} = 28 \cr} $$