Question

If $x$ is real, $x+\frac{1}{x}\ne 0$ and $x^3+\frac{1}{x^3}=0$, then the value of ${(x+\frac{1}{x})}^4$ is

• A. $4$
• B. $9$
• C. $16$
• D. $25$

Answer 1

The correct option is B $9$

Given that, $x^3+\frac{1}{x^3}=0$
We know that,
${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}=3(x+\frac{1}{x})\Rightarrow {(x+\frac{1}{x})}^3=0+3(x+\frac{1}{x}){(x+\frac{1}{x})}^2=3\Rightarrow {\left[{(x+\frac{1}{x})}^2\right]}^2=(3{)}^2=9$.