If $p,q,r$ are the zeros of the polynomial $x^3-3px^2+3q^2x-r^3$ then show that $p=q=r$. If $p,q,r$ are the zeros of the polynomial $x^3-3px^2+3q^2x-r^3$ then show that $p=q=r$.
I've tried using Vieta's theorem but still have not found any result. I feel that it may be solved using Vieta's theorem. Or, is there any other way? 

Per Nate's and zyx's comments, the base field is assumed to have characteristic not equal to $2$, $3$, or $5$. Let $$f(x):=x^3-3px^2+3q^2x-r^3\,.$$  We have $f(p)=f(q)=f(r)=0$.  It is easy (but not trivial, and this part involves divisions by $2$, $3$, and $5$) to see that $r=0$ implies that $p=q=0$.  From now on, it is assumed that $r\neq 0$.  
Write $u:=\frac{p}{r}$ and $v:=\frac{q}{r}$.  Then, $f(r)=0$ implies that (with divison by $3$)$$v^2=u\,.\tag{1}$$  Plugging in this result into $f(q)=0$, one gets  $$v^2\left(4v-3v^2\right)=u(4v-3u)=1\,.\tag{2}$$
That is,
$$(v-1)^2\left(3v^2+2v+1\right)=3v^4-4v^3+1=0\,.\tag{3}$$
Noe, $f(p)=0$ yields
$$u\left(3v^2-2u^2\right)=1\,.\tag{4}$$
From (2) and (4), one obtains $u\neq 0$, whence $v\neq 0$, by (1).  From (2) and (4), it follows that
$$3u-2u^2=3v^2-2u^2=4v-3u$$
or, due to (1) (with division by $2$),
$$v^4=u^2=3u-2v=3v^2-2v\,.\tag{5}$$
As $v\neq 0$, (5) becomes
$$(v-1)^2(v+2)=v^3-3v+2=0\,.\tag{6}$$
The only solution to (3) and (6) is $v=1$.  Thus, $u=v=1$, or $p=q=r$.
