Answer
See proof below.
Work Step by Step
If a cubic polynomial $f(x)$ had more than three zeros, we observe four of them $z_{1},\ z_{2},\ z_{3}$ and $z_{4}$, and apply Rolle's Theorem:
over $[z_{1},z_{2}] \Rightarrow$ there exists a $c_{1}\in(z_{1},z_{2})$ such that $f'(c_{1})=0$.
Applying the theorem over $[z_{2},z_{3}]$ and $[z_{3},z_{4}]$, we find that there exist antoher two zeros of $f'$. Call these $c_{2}$ and $c_{3}$.
Now, $c_{1},c_{2}$ and $c_{3}$ are zeros of $f'$. Rolles theorem guarantees two zeros of $f''$,
$d_{1}\in(c_{1},c_{2})$ and $d_{2}\in(c_{2},c_{3}).$
Next, with $d_{1}$ and $d_{2}$ being zeros of $f''$, Rolle's theorem guarantees that
$f''' $has a zero in $(d_1,d_2)$.
But
$f(x)=ax^{3}+bx^{2}+cx+d$
$f'(x)=3ax^{2}+3bx+c$
$f''(x)=6ax+3b$
$f'''(x)=6a$
$f'''$ has no zeros ( a is not zero because f has degree 3). The initial assumption leads to a contradiction.
Therefore, $f(x)$ can not have more than three zeros.
