10. For what values of the parameter $a$ does the equation $x^{4}-40 x^{2}+144=a\left(x^{2}+4 x-12\right)$ have exactly three distinct solutions?
Solution: Factorize the right and left sides of the equation:

$(x-2)(x+2)(x-6)(x+6)=a(x-2)(x+6)$. The equation can be written as
$(x-2)(x+6)\left(x^{2}-4 x-12-a\right)=0$. It is obvious that 2 and -6 are roots of this equation. We are satisfied with the situation where exactly one of the roots of the equation $x^{2}-4 x-12-a=0$ does not coincide with 2 and -6. The total number of solutions will be three if the last equation has one root $x_{3}$, such that $x_{3} \neq x_{1}, x_{3} \neq x_{2}$, or if this equation has two roots $x_{3}, x_{4}$, one of which coincides with $x_{1}$ or $x_{2}$.

The equation $x^{2}-4 x-12-a=0$ has one root when $a=-16$ and this root coincides with 2. This case does not suit us. The number 2 is a root of this equation when $a=-16$. This case has already been considered. The number -6 is a root of this equation when $a=48$, and in this case, the second root is different from 2 and -6 - Answer: $a=48$.