2. Find all real solutions to the equation
$$
\left(x^{2}+3 x+1\right)^{x^{2}-x-6}=1 \text {. }
$$
Solution: Let $a=x^{2}+3 x+1$ and let $b=x^{2}-x-6$. The only way to have $a^{b}=1$, is if $a= \pm 1$ or $b=0$.
- If $b=0$, then we solve the quadratic $x^{2}-x-6=0$ which has solutions $x=-2,3$. (we would also have to check that $a \neq 0$ in this case)
- If $a=1$, then we solve the quadratic $x^{2}+3 x+1=1$ which has solutions $x=0,-3$.
- If $a=-1$, then we solve the quadratic $x^{2}+3 x+1=-1$ which has solutions $x=-1,-2$. (we also have to check that $b$ is an even integer in this case)

Therefore there are a total of 5 candidate solutions: $x=-3,-2,-1,0,3$.
Remark: In order to receive full marks, a student would have to demonstrate that $x=-3,-2,-1,0,3$ are actually all solutions, by substituting each of these values into the expression, and verify that the result is indeed 1 .