For what values of a there are exactly two real numbers that satify the equation:
x^3 + (a-1)x^3 + (9-a)x - 9 = 0 ?
x = {$\displaystyle 1$, $\displaystyle \frac{1}{2}\,{\frac {-a+\sqrt {{a}^{2}-36\,a}}{a}}$, $\displaystyle \frac{1}{2}\,{\frac {-a-\sqrt {{a}^{2}-36\,a}}{a}}$}
So, 1 is a real number. Now pick an 'a' such that the 2nd and 3rd element are equal.
$\displaystyle a = 36$
Note: I only checked the 'a' in the real. There might be other ways of satisfying the question.
$\displaystyle x = \{ 1 , -\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} , -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36} \}$
Make the 2nd and 3rd element equal:
$\displaystyle -\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} = -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36}$
$\displaystyle a = \{6, -6\} $