Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Correct Answer: $\frac{3\sqrt{15}}{8}$
Solution :
Given: $x^2+\frac{1}{x^2}=\frac{7}{4}$
⇒ $x^2+\frac{1}{x^2}+2=\frac{7}{4}+2$ [adding 2 on both sides]
⇒ $x^2+\frac{1}{x^2}+2×x^2×\frac{1}{x^2}=\frac{15}{4}$
⇒ $(x+\frac{1}{x})^2=\frac{15}{4}$
$\therefore (x+\frac{1}{x})=\frac{\sqrt{15}}{2}$
Now, we know,
$(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})$
⇒ $(\frac{\sqrt{15}}{2})^3=x^3+\frac{1}{x^3}+3×\frac{\sqrt{15}}{2}$
$\therefore x^3+\frac{1}{x^3}= \frac{15\sqrt{15}}{8}-\frac{3\sqrt{15}}{2}=\frac{3\sqrt{15}}{8}$
Hence, the correct answer is $\frac{3\sqrt{15}}{8}$.
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