Question : If $(x+\frac{1}{x})\neq 0$ and $(x^3+\frac{1}{x^3})= 0$, then the value $(x+\frac{1}{x})^4$ is:
Option 1: 9
Option 2: 12
Option 3: 15
Option 4: 16
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Correct Answer: 9
Solution :
Given: $(x+\frac{1}{x})\neq 0$ and $(x^3+\frac{1}{x^3})=0$
$(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})$
Substitute the value of $(x^3+\frac{1}{x^3})=0$ in above equation, we get,
⇒ $(x+\frac{1}{x})^3=0+3(x+\frac{1}{x})$
⇒ $(x+\frac{1}{x})^3=3(x+\frac{1}{x})$
⇒ $(x+\frac{1}{x})^{3–1}=3$
⇒ $(x+\frac{1}{x})^2=3$
On squaring both sides of the above equation, we get,
⇒ $((x+\frac{1}{x})^2)^2=3^2$
⇒ $(x+\frac{1}{x})^4=9$
Hence, the correct answer is 9.
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