Common roots of two polynomials

Reduce $P$ modulo $Q$. From $Q(\alpha)=0$: $\alpha^2=-\alpha-a$. Then
$$\alpha^3=\alpha\cdot\alpha^2=\alpha(-\alpha-a)=-\alpha^2-a\alpha=-(-\alpha-a)-a\alpha=\alpha+a-a\alpha=(1-a)\alpha+a.$$
So $P(\alpha)=\alpha^3+a\alpha+1=(1-a)\alpha+a+a\alpha+1=(1-a+a)\alpha+(a+1)=\alpha+a+1$.
Thus $P(\alpha)=0\iff\alpha=-a-1$.

Substitute into $Q$. $Q(-a-1)=(-a-1)^2+(-a-1)+a=a^2+2a+1-a-1+a=a^2+2a=a(a+2)$.

Conclusion. $a\in\{0,-2\}$.

Check. For $a=0$: $P=X^3+1=(X+1)(X^2-X+1)$, $Q=X^2+X=X(X+1)$. Common root $-1$. For $a=-2$: $P=X^3-2X+1$, $Q=X^2+X-2=(X-1)(X+2)$. $P(1)=0$, common root $1$. Correct.