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? 

Using the relation between the roots of a polynomial and the coefficients of its terms, we get that 
$$p+q+r=3p\tag1$$ $$pq+qr+pr=3q^2\tag2$$ $$pqr=r^3\tag3$$
Now $(1)$ gives $q+r=2p$ and $(3)$ gives $pq=r^2\tag4$ assuming $r\not=0$.
If $r=0$, we can write that $q=2p$ and $p=3q$ which means $p=q=r=0$. 
So from $(2)$, we can say now that $$r^2+qr+pr=3q^2$$ $$\implies r(r+q)+pr=3q^2$$ $$\implies 2pr+pr=3q^2$$ $$\implies pr=q^2\tag5$$
$(4)$ divided by $(5)$ gives $\frac{pq}{pr}=\frac{r^2}{q^2} \implies q^3=r^3 \implies q=r$ assuming that $p \not=0$. If $p=0$, then it can be similarly proved that $p=q=r=0$.
Similarly, we can also prove that $p=q$.
Hope this helps.
