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