Question : If $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$
Option 1: $36 \sqrt{5}$
Option 2: $40 \sqrt{5}$
Option 3: $46 \sqrt{5}$
Option 4: $34 \sqrt{5}$
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Correct Answer: $34 \sqrt{5}$
Solution :
$x^{2} + \frac{1}{x^{2}} = 18$
⇒ $x^{2} + \frac{1}{x^{2}} + 2 = 18+2$
⇒ $(x + \frac{1}{x})^{2} = 20$
⇒ $(x + \frac{1}{x}) = \sqrt{20}$
Now, $(x + \frac{1}{x})^{3}=x^3+\frac{1}{x^3}+3 × x × \frac{1}{x} × (x + \frac{1}{x})$
⇒ $(\sqrt{20})^3=x^3+\frac{1}{x^3}+3×\sqrt{20}$
⇒ $x^3+\frac{1}{x^3}=20\sqrt{20} - 3\sqrt{20}$
⇒ $x^3+\frac{1}{x^3}=17\sqrt{20}$
$\therefore x^3+\frac{1}{x^3}=34\sqrt{5}$
Hence, the correct answer is $34\sqrt{5}$.
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