Let the common root be $t$. Adding the three equations, we get
$$(a + b + c)(t^3 + t + 1) = 0.$$
If $t^3 + t + 1 = 0$, then $at^3 + bt + c = 0$ becomes $(b-a)t + (c-a) = 0$. Similarly, $bt^3 + ct + a = 0$ becomes $(c-b)t + (a-b) = 0$. As $a$, $b$, $c$ are distinct, we obtain $\frac{c-a}{a-b} = t = \frac{a-b}{b-c}$. After expansion and regrouping the terms, this becomes
$$(a-b)^2 + (b-c)^2 + (c-a)^2 = 0.$$
This is a contradiction since $a$, $b$, $c$ are distinct.
Therefore, we must have $a+b+c=0$. In that case, it is obvious that $t=1$ is the common root. As $a$, $b$, $c$ are nonzero, there are two possible cases.
If two of $a$, $b$, $c$ are positive, say $a$, $b > 0$, we consider $f(x) = bx^3 + cx + a$. Since $f(0) = a > 0$ and $f(x) < 0$ when $x \to -\infty$, there is at least one negative root of $f(x)$. As $1$ is another root and a cubic polynomial cannot have exactly two real roots, the equation $f(x) = 0$ must have exactly three real roots.
Similarly, if two of $a$, $b$, $c$ are negative, say $a$, $b < 0$, then the above argument shows $g(x) = -(bx^3 + cx + a)$ has three real roots. This shows $f(x) = 0$ also has three real roots.

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

If $t^3 + t + 1 = 0$, then $at^3 + bt + c = 0$ becomes $(b - a)t + (c - a) = 0$. Similarly, $bt^3 + ct + a$ 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 $f(y) = b y^3 + c y + a$. 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 $g(y) = -(b y^3 + c y + a)$ has three real zeros, and the same is clearly true for $f(x)$.