If $${x^2} + \frac{1}{{{x^2}}} = 98{\text{,}}$$   $$\left( {x > 0} \right){\text{,}}$$   then the value of $$x^3 + \frac{1}{{{x^3}}}$$  is?

Solution(By Examveda Team)

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 98 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 100 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 100 \cr & \Rightarrow x + \frac{1}{x} = 10 \cr & {\text{Cubing}}\,{\text{both}}\,{\text{sides}} \cr & \Rightarrow {\text{ }}{x^3} + \frac{1}{{{x^3}}} = {10^3} - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 1000 - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 970 \cr} $$