Find the intervals in which the function $f$ given by $f(x) = x^{3} + \frac{1}{x^{3}}, x \neq 0$ is:
$(i)$ increasing
$(ii)$ decreasing.

(N/A) $f(x) = x^{3} + x^{-3}$
$\therefore f'(x) = 3x^{2} - 3x^{-4} = 3x^{2} - \frac{3}{x^{4}} = \frac{3(x^{6} - 1)}{x^{4}}$
To find the critical points,set $f'(x) = 0$:
$3(x^{6} - 1) = 0 \Rightarrow x^{6} = 1 \Rightarrow x = \pm 1$
Since $x \neq 0$,the points $x = -1, 0, 1$ divide the real line into intervals $(-\infty, -1), (-1, 0), (0, 1), (1, \infty)$.
For $x \in (-\infty, -1)$,$f'(x) > 0$,so $f$ is increasing.
For $x \in (-1, 0)$,$f'(x) < 0$,so $f$ is decreasing.
For $x \in (0, 1)$,$f'(x) < 0$,so $f$ is decreasing.
For $x \in (1, \infty)$,$f'(x) > 0$,so $f$ is increasing.
Thus,$f$ is increasing in $(-\infty, -1) \cup (1, \infty)$ and decreasing in $(-1, 0) \cup (0, 1)$.