If $$\frac{{{x^2} + 1}}{x} = 4\frac{1}{4},$$   then what is the value of $${x^3} + \frac{1}{{{x^3}}}?$$

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{x^2} + 1}}{x} = 4\frac{1}{4} \cr & x + \frac{1}{x} = \frac{{17}}{4} \cr & {\text{On cubing both sides,}} \cr & {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\frac{{17}}{4}} \right)^3} \cr & {x^3} + \frac{1}{{{x^3}}} + 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = \frac{{4913}}{{64}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3\left( {\frac{{17}}{4}} \right) = \frac{{4913}}{{64}} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{4913}}{{64}} - \frac{{51}}{4} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{4097}}{{64}} \cr} $$