CHKMO 2015 and cubic equations Let $a,b,c$ be distinct real numbers. If the equations $E_1: ax^3+bx+c=0, E_2: bx^3+cx+a=0$ and $E_3: cx^3+ax+b=0$ have a common root, prove that at least one of these equations has three real roots(not necessarily distinct).
Suppose $r$ is a common root, then adding the three equations implies $r$ is also a root of $(a+b+c)x^3+(a+b+c)x+(a+b+c)=0$. If $a+b+c\not=0,$ then $r$ is a root of $x^3+x+1=0$, which has one real root and two complex roots. If $r$ is complex, then we can easily derive a contradiction by using viete theorem, hence $r$ is real. Further more observation shows that if $p_i$ is a complex root of $E_i$, then $Re (p_i)=Re (p_j)$. But then I can't really move on, please helps.
 A: If $abc = 0$, WLOG $a = 0$, and so the common root must b$ x = - \frac{c}{b}$. However, it is easy to check that $ 0 = b \frac{-c^3}{b^2} + c \frac{-c}{b} = \frac{ -c^3-c^2b } { b^2} = - \frac{ c^2 (c+b) } {b^2}$, hence we must have $ c = -b$, which gives us $x=1$. Then, $ 0 = bx^3 + cx + a = bx^3 - bx = bx(x-1)(x+1)$ would have 3 real roots and we are done. Henceforth, $abc \neq 0$.
If $ a+b+c\neq 0$, then $r^3+r+1 = 0$ and $ r\neq 0$. WLOG, let $c = \min(a,b,c)$, and form the first equation we get $ (a-c)r^3 + (b-c)r = 0 \Rightarrow (a-c)r^2 + (b-c) = 0 $. However, since $ a-c > 0, b -c > 0$, we must have $ (a-c)r^+ (b-c) > 0  $ which is a contradiction. hence we must have $a+b+c=0$.
Thus, $r= 1$ is a root. Observe that $(ax^3 + bx+c) / (x-1) = ax^2 + ax + a + b $. For this to have 2 real roots, it is necessary and sufficient that $ \frac{ -b} {a}\geq \frac{3}{4}$. 
WLOG, $ab>0$.  WLOG, $a, b,-c > 0 $. We want to show that we have either $ \frac{-c}{b} \geq \frac{3}{4} $ or $ \frac{a}{-c} \geq \frac{3}{4}$ or $\frac{-b}{a} \geq \frac{3}{4}$. Suppose not, then we have $ 4 (-c) < 3b, \ldots$. However, summing these up we get $ a + b + c < 0$, which is a contradiction.
Hence, one of these inequalities must hold and the corresponding equation has 3 real roots.
