Answer
$x=-3;x=-1;x=1$;
Work Step by Step
We use a graphing calculator to graph the function $y=x^3+3x^2-x-3$.
From the graph we notice that the $x$-intercepts are:
$$\begin{align*}
x_1&=-3\\
x_2&=-1\\
x_3&=1.
\end{align*}$$
We check if the values $x_1$ and $x_2$ check the equation $x^3+3x^2-x-3=0$:
$$\begin{align*}
x_1&=-3\\
(-3)^3+3(-3)^2-(-3)-3&\stackrel{?}{=}0\\
-27+27+3-3&\stackrel{?}{=}0\\
0&=0\checkmark\\\\
x_2&=-1\\
(-1)^3+3(-1)^2-(-1)-3&\stackrel{?}{=}0\\
-1+3+1-3&\stackrel{?}{=}0\\
0&=0\checkmark\\\\
x_3&=1\\
1^3+3(1^2)-1-3&\stackrel{?}{=}0\\
1+3-1-3&\stackrel{?}{=}0\\
0&=0\checkmark
\end{align*}$$
Therefore the solutions of the equation are $-3$, $-1$ and $1$.
