If $p$ is a polynomial of degree 3, then between any two consecutive zeroes of p, can there be two zeroes of its derivative ( $p'$) ?
Suppose if there are two zeroes of $p'$ between two consecutive zeroes of $p$, then the graph needs to have structure which looks like the pic below
From the graph, it is clear that $p'$ has $3$ zeroes, which is not possible for cubic polynomial.
I tried to prove this by using Rolle's theorem. I was only successful in showing that cubic polynomial has no constant term. I would like to know if there is any other way to prove this apart from graphical solution.
If by "two roots between" you mean two distinct roots strictly between two roots of $p(x)$, you are correct: $p'(x)$ cannot have two distinct roots between consecutive roots of $p(x)$, if $p(x)$ is a cubic polynomial with real coefficients.
Say two roots of $p(x)$ are $x=a$ and $x=b$, $a\lt b$, and the two roots of $p'(x)$ are $r_1$ and $r_2$, with $a\lt r_1\lt r_2\lt b$. We will show that $p(x)$ has a zero strictly between $a$ and $b$, so that they are not "consecutive roots".
Since $p'(x)$ is a quadratic, it follows that $p'(x)$ changes signs at $x=r_1$ and at $x=r_2$. Replacing $p(x)$ with $-p(x)$ if necessary, we may assume that $p'(x)$ is positive to the left of $r_1$, negative between $r_1$ and $r_2$, and positive to the right of $r_2$.
That means that $p(x)$ is increasing on $(-\infty,r_1)$, decreasing on $(r_1,r_2)$, and increasing on $(r_2,\infty)$. Thus, $p(r_1)\gt 0$ (since $p(a)=0$ and $a\lt r_1$); and $p(r_2)\lt 0$ (because $p(b)=0$ and $r_2\lt b$). By the intermediate value theorem, it follows that $p(x)$ has a zero between $r_1$ and $r_2$, hence $p(x)$ has a zero between $a$ and $b$.
On the other hand, if you allow one of the roots of $p'(x)$ to coincide with either $a$ or $b$, then you can take $a=0$, $b=1$, and $p(x) = x(x-1)^2$ as a counterexample; here you get $p'(x) = (x-1)^2 + x(x-1) = (x-1)(x-1+x) = (x-1)(2x-1)$, which has roots at $x=\frac{1}{2}$ and $x=1$, both "between" $a$ and $b$ (since we are allowing them to equal $b$ in this scenario).
If you allow the two roots of $p'(x)$ to coincide, that is, to be a double root, then you are fine again unless all the roots for both $p(x)$ and $p'(x)$ coincide. By doing a horizontal shift we may assume that the double root of $p'(x)$ is at $0$, so that $p'(x) = Ax^2$ for some nonzero $A$; then $p(x)=ax^3+c$ for some $c$, and this polynomial has a single real root at $x=\sqrt[3]{-\frac{c}{a}}$; but if $c\neq 0$ then the roots of $p'(x)$ are not "between" the roots of $p(x)$. And if $c=0$, then all roots are the same, which of course is possible.