Find the intervals in which the function

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

$\therefore f^{\prime}(x)=3 x^{2}-\frac{3}{x^{4}}=\frac{3 x^{6}-3}{x^{4}}$

Then, $f^{\prime}(x)=0 \Rightarrow 3 x^{6}-3=0 \Rightarrow x^{6}=1 \Rightarrow x=\pm 1$

Now, the points $x=1$ and $x=-1$ divide the real line into three disjoint intervals i.e., $(-\infty,-1),(-1,1)$, and $(1, \infty)$.

In intervals $(-\infty,-1)$ and $(1, \infty)$ i.e., when $x<-1$ and $x>1, f^{\prime}(x)>0$.

Thus, when $x<-1$ and $x>1, f$ is increasing.

In interval $(-1,1)$ i.e., when $-1

Thus, when $-1