If $p,q,r$ are the zeros of the polynomial $x^3-3px^2+3q^2x-r^3$ then show that $p=q=r$.

Question

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?

Answer 1

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.

Answer 2

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 $p=q=r$.

Answer 3

If the numbers are real, then this is true independent of the middle coefficient (if the numbers are nonzero), and the exceptional case where $r=0$ does not present any difficulty.

$p$ is the arithmetic mean, and $r$ the geometric mean, of the three numbers $p,q,r$. At least one of $p$ and $r$ must be extremal in the sense of being the largest or smallest of the numbers. If the arithmetic mean $p$ is extremal then all the numbers must be equal. If the geometric mean $r$ is extremal and nonzero then all the numbers must be equal.

If $r=0$, the equations become $p+q=3p$ and $pq=3q^2$, which leads to $p=q=0$ by eliminating either variable.

Answer 4

Here is a conceptual and generalizable solution, valid over the complex numbers.

Each of $|p|,|q|,|r|$ is equal to some function of $(p,q,r)$, depending on all of its arguments (so $F(p,q,r)=|p|$ does not work, but $|p|=|\frac{p+q+r}{3}|$ does), with the property that:

1. the function is no larger in absolute value than the maximum absolute value of its arguments
2. if the function equals the maximum absolute value of its arguments, then all the arguments have the same absolute value

Because at least one number in the set has to have the largest absolute value, properties (1) and (2) mean that they all have the same absolute value. Now we need one further fact: at least one of the functions computing $|p|,|q|, ...$ should have the stronger version of property (2), that

2.+ : if |function| equals the maximum absolute value of its arguments, then all the arguments are equal.

In the current problem, the stronger property $(2+)$ is true for $|p|$ = |arithmetic mean|, and for the more complicated expression for $|q|$, but not for $|r|$ which is the geometric mean of absolute values and does not depend on the phases of the numbers.

This immediately implies the degree $n$ generalization of the present problem, but it can also be applied to any system of equations (not only from Vieta's formulas, and not necessarily polynomial) where the absolute values of the numbers are some kind of generalized mean.

It also is not necessary for all the absolute value of |p|, etc to be functions of the other numbers, only bounded by functions that have the above properties.