10. For what values of the parameter $a$ does the equation $x^{4}-20 x^{2}+64=a\left(x^{2}+6 x+8\right)$ have exactly three distinct solutions?
Solution: Factorize the right and left sides of the equation:

\((x-2)(x+2)(x-4)(x+4)=a(x+2)(x+4)\). The equation can be written as \((x+2)(x+4)\left(x^{2}-6 x+8-a\right)=0\). It is obvious that -2 and -4 are roots of this equation. We are satisfied with the situation where exactly one of the roots of the equation \(x^{2}-6 x+8-a=0\) does not coincide with -2 and -4. The total number of solutions will be three if the last equation has one root \(x_{3}\), and \(x_{3} \neq x_{1}, x_{3} \neq x_{2}\), or if this equation has two roots \(x_{3}, x_{4}\), one of which coincides with \(x_{1}\) or \(x_{2}\). The equation \(x^{2}-6 x+8-a=0\) has one root when \(a=-1\) and this root is different from the specified numbers. The numbers -2 and -4 are roots of the last equation when \(a=24\) and \(a=48\), and in these cases, the second roots are different from -2 and -4. Answer: \(a=-1 ; 24 ; 48\).

## VARIANT 4