Bulgarian Mathematical Olympiad

Solution:
The first equation is quadratic with discriminant $D=(9 a^{2}-1)^{2}$. Therefore it has two different solutions for $a \neq \pm \frac{1}{3}$ and exactly one solution for $a= \pm \frac{1}{3}$.

Since the function
$$2 x^{3}+6 x+(3 a-1)^{2} 12^{x}$$
is strictly increasing, the second equation has at most one solution. For $a=\frac{1}{3}$ it has solution $x=0$, while for $a=-\frac{1}{3}$ it is not defined. Finally, the only value of $a$ which satisfies the condition, is $a=\frac{1}{3}$.