The Seventeenth Hong Kong (China) Mathematical Olympiad

Let $t$ be the common root. Adding the three equations $a{t}^3+bt+c=0$, $b{t}^3+ct+a=0$ and $c{t}^3+at+b=0$, we get $(a+b+c)(t^3+t+1)=0$.

If $t^3+t+1=0$, then $a{t}^3+bt+c=0$ becomes $(b-a)t+(c-a)=0$. Similarly, $b{t}^3+ct+a$ becomes $(c-b)t+(a-b)=0$. As $a$, $b$, $c$ are distinct, the two equations combine to give $\frac{c-a}{b-a}=\frac{a-b}{c-b}$, i.e. $(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$, which yields the contradiction $a=b=c$.

Therefore, we must have $a+b+c=0$. Then it is obvious that $t=1$ is the common root. As $a$, $b$, $c$ are nonzero, there are two possible cases.

Case 1: Two of $a$, $b$, $c$ are positive, say $a$ and $b$. Consider $f(y)=b{y}^3+cy+a$. Since $f(0)=a>0$ and $f$ is negative for sufficiently small $y$, there is at least one negative root of $f(x)=0$. As $1$ is also a root, $f(x)=0$ has at least two real roots and hence exactly three real roots (since complex roots occur in pairs).

Case 2: Two of $a$, $b$, $c$ are negative, say $a$ and $b$. Then the same argument shows that the polynomial $g(y)=-(b{y}^3+cy+a)$ has three real zeros, and the same is clearly true for $f(x)$.