Find the intervals in which the function $f(x) = (x+1)^{3}(x-3)^{3}$ is strictly increasing or strictly decreasing.

(A) Given function is $f(x) = (x+1)^{3}(x-3)^{3}$.
First,we find the derivative $f'(x)$ using the product rule:
$f'(x) = 3(x+1)^{2}(x-3)^{3} + 3(x-3)^{2}(x+1)^{3}$
$f'(x) = 3(x+1)^{2}(x-3)^{2} [(x-3) + (x+1)]$
$f'(x) = 3(x+1)^{2}(x-3)^{2} (2x-2)$
$f'(x) = 6(x+1)^{2}(x-3)^{2}(x-1)$
To find the critical points,set $f'(x) = 0$:
$6(x+1)^{2}(x-3)^{2}(x-1) = 0$
This gives $x = -1, 1, 3$.
These points divide the real line into four intervals: $(-\infty, -1)$,$(-1, 1)$,$(1, 3)$,and $(3, \infty)$.
$1$. For $x \in (-\infty, -1)$,$f'(x) < 0$,so $f$ is strictly decreasing.
$2$. For $x \in (-1, 1)$,$f'(x) < 0$,so $f$ is strictly decreasing.
$3$. For $x \in (1, 3)$,$f'(x) > 0$,so $f$ is strictly increasing.
$4$. For $x \in (3, \infty)$,$f'(x) > 0$,so $f$ is strictly increasing.
Thus,$f$ is strictly decreasing in $(-\infty, -1) \cup (-1, 1)$ and strictly increasing in $(1, 3) \cup (3, \infty)$.