InTeRvAl

Not complete :(... How about this?

We are given that $a$ is a reciprocal of any number, so let $a=\frac{1}{k}$ . Then, we equate 0 with the function $x^2(3a-x)\implies x^2(\frac{3}{k}-x)$ .

$0=x^2(\frac{3}{k}-x)$

$0=\frac{3}{k} - x$

$x=\frac{3}{k}$

Okay, so now, we can deduce that if $k$ is an integer, the values that satisfy $x$ are $1, -1$ . This implies to that no matter what $3a-x$ is 0. Therefore, the value of $f(0)$ is $\boxed0$ . Q.E.D.