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

$\left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)=3\left(x+\frac{1}{x}\right)$

$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=3$

$\therefore\left(x+\frac{1}{x}\right)^{4}=3 \times 3=9$