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If $${x^2} + \frac{1}{{{x^2}}} = 7,$$ then the value of $${x^3} + \frac{1}{{{x^3}}}$$ where x > 0 is equal to:
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& {x^2} + \frac{1}{{{x^2}}} = 7 \cr
& x + \frac{1}{x} = 3 \cr
& {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 = 18 \cr} $$
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