Irish Mathematical Olympiad

Reduce the given polynomial to normal form, which can be done by noting that
$$\begin{aligned} x^3 - 3a x^2 + 2x - 4a &= (x-a)^3 - 3a^2 x + a^3 + 2(x-a) - 2a \\ &= (x-a)^3 + (2 - 3a^2)(x-a) - 2a^3 - 2a \\ &= X^3 + pX + q, \end{aligned}$$
where $X = x - a$, and $p = (2 - 3a^2)$, $q = -2a(a^2 + 1)$. The condition that this cubic has three real roots is that
$$\frac{(2 - 3a^2)^3}{27} + \frac{(2a(a^2 + 1))^2}{4} = \frac{p^3}{27} + \frac{q^2}{4} < 0.$$
Letting $t = a^2$ the requirement is that $(2 - 3t)^3 + 27t(t + 1)^2 < 0$, i.e., that $108t^2 - 9t + 8 < 0$. But $108t^2 - 9t + 8 = 3(6t - \frac{1}{4})^2 + 8 - \frac{3}{16}$ and so there is no real number $t$ for which this is negative. Therefore, the cubic cannot have three real roots. As its coefficients are real numbers, this implies that it has one and only one real root.