Find value of $b$ and $c$ Suppose $x^{12} - 1$ is divisible by $x^4 + bx^2 + c$ . Find all possible values for $b$ and $c$ . I've tried to use $x^4 + bx^2 + c = 0$ for finding remainder and equaling it with zero but didn't help for finding $b$ and $c$. 

$$x^{12}-1=(x^6-1)(x^6+1)=(x^2-1)(x^4+x^2+1)(x^2+1)(x^4-x^2+1)=(x^4-1)(x^4+x^2+1)(x^4-x^2+1)$$
The situation is a bit trickier since those $3$ don't need to be the only possible solutions. Using the substitution suggested in comments by MathematicianByMistake $t=x^2$ we get $t^6-1=(t^3-1)(t^3+1)=(t-1)(t+1)(t^2+t+1)(t^2-t+1)$ needs to be divisible by $t^2+bt+c$, since $t^2-t+1$ and $t^2+t+1$ are irreducible you're left with $(t-1)(t+1)$ which can be combined only in one way to get $t^2-1$ hence those are the only $3$ choices.
