Find all real numbers a such that the equation x3−ax2−2ax+a2−1=0 has exactly one real solution in x.
a2−1=(a+1)(a−1)
so the most likely possibilities are
[x−(a+1)]3[x+(a+1)]3[x−(a−1)]3[x+(a+1)]3
I'd eliminate the last one becasue there won't be any minus signs.
I'd test the others.
+99 
