If $ax^3+bx+c=0$, $bx^3+cx+a=0$, $cx^3+ax+b=0$ have a common root for distinct non-zero $a$, $b$, $c$, then at least one has three real roots 
Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations
$$\begin{align}
E_1: ax^3+bx+c=0 \\
E_2: bx^3+cx+a=0 \\
E_3: cx^3+ax+b=0
\end{align}$$
have a common root, prove that at least one of these equations has three real roots (not necessarily distinct).

This question was a math olympiad problem.Below was my attempt but I've got confused:
I know that the discriminant of the equation $x^3+Ax+B=0$ is given by $-4A^3-27B^2$, so it suffices to prove that the discriminant of at one least of the equations is non-negative.
It's easy to prove that the common root $r$ must be real. Now assume on the contrary that all three equations had a pair of complex conjugate roots. Let $\alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma, \bar{\gamma}$ be the complex conjugate roots of $E_1, E_2$ and $E_3$ respectively.
Then by viete theorem, we have $$\alpha \bar{\alpha} r=|\alpha|^2r=\frac{-c}{a},$$
$$|\beta|^2r=\frac{-a}{b},$$
$$|\gamma|^2r=\frac{-b}{c}.$$
Multiplying the three equations gives $$|\alpha \beta \gamma|^2r^3=-1.$$
Hence $r<0.$ Hence $-c/a<0$ or $c/a>0.$ Similarly $a/b>0$ and $b/c>0.$ This implies that $a,b,c$ all all positive or or negative.
By symmetry, we can assume $a>b>c>0$, but then none of the discriminant could be non-negative. So what's wrong with the argument and how can I complete the proof? Thanks in advance!
 A: Let $r$ be the common root, then
$$(a+b+c)(r^3+r+1)=0 \tag{$E_1+E_2+E_3$}$$
Case I:

If $r^3+r+1=0$, then $(ar^3+br+c)-a(r^3+r+1)=0$
$$\implies (a-b)r=(c-a)$$
By symmetry, $$r=\frac{c-a}{a-b}=\frac{a-b}{b-c}=\frac{b-c}{c-a}$$
$$c=\frac{a+b \pm i(a-b)\sqrt{3}}{2}$$
Or equivalently,
$$r=\frac{-1 \pm i\sqrt{3}}{2}$$
that contradicts with $r^3+r+1=0$.

Case II:

If $a+b+c=0$, then $r=1$.
For $x\neq 1$,
$$\frac{ax^3+bx+c}{x-1}=ax^2+ax-c=0$$
$$\Delta_{1}=a^2+4ac$$
Similarly,
$$\Delta_{2}=b^2+4ba$$
$$\Delta_{3}=c^2+4cb$$
Since $a+b+c=0$, either two of $a$, $b$ and $c$ are of equal sign.
That is either one of $ab$, $bc$ and $ca$ is positive.
At least one of the $\Delta_{1}$, $\Delta_{2}$ and $\Delta_{3}$ is positive.


Further points to be noticed
For $r \ne 1$ and $a+b+c=0$,
$$\frac{r^3-1}{r-1}=-\frac{b}{a}=-\frac{c}{b}=-\frac{a}{c} \implies a=b\omega=c\omega^2 \land \omega^3=1$$
which does not give distinct, non-zero and real values of $a$, $b$ and $c$.
Also considering the matrix equation:
$$
\begin{pmatrix}
  a & b & c \\
  b & c & a \\
  c & a & b
\end{pmatrix}
\begin{pmatrix}
  r^3 \\  r \\  1
\end{pmatrix}=
\begin{pmatrix}
  0 \\  0 \\  0
\end{pmatrix}$$
The eigenvalues are $\lambda_1=a+b+c$ and
$\lambda_{2,3}=\pm \sqrt{\frac{(b-c)^2+(c-a)^2+(a-b)^2}{2}}$.
For the matrix is singular,
$$\lambda_1 \lambda_2 \lambda_3=0$$
The eigenvector corresponding to $\lambda_1=a+b+c$ is
$$\vec{v_1}=(1,1,1) \implies r^3=r=1$$
For $a=b=c$, $$\lambda_{2,3}=0 \implies r^3+r+1=0$$
A: Suppose all three have only one solution.
Then $3ax^2+b$ is always positive or negative i.e. $ab<0$. The same way we get $ac<0$ and $bc<0$. Impossible.
