Question : If $(x+\frac{1}{x})^{2}=3$, then the value of $(x^{3}+\frac{1}{x^{3}})$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
Correct Answer: 0
Solution :
Given: $(x+\frac{1}{x})^{2}=3$,
Taking square root on both sides, we get
$(x+\frac{1}{x})=\sqrt{3}$
Now, Cubing both sides we get
$(x+\frac{1}{x})^3=(\sqrt{3})^3$
⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=(3\sqrt{3})$
⇒ $x^3+\frac{1}{x^3}=3\sqrt{3}–3(x+\frac{1}{x})$
$\because x+\frac{1}{x}=\sqrt{3}$
Thus, $x^3+\frac{1}{x^3}=3\sqrt{3}–3\sqrt{3} = 0$
Hence, the correct answer is 0.
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