Call $Q_1=x^3+ax^2+1, Q_2=x^3-ax^2+1$ and assume $\alpha\neq0$. First one can easily prove that every root of $Q_1$ has a root of $Q_2$ such that the difference of the $2$ roots is an integer (or $P\equiv 0$). Let the roots of $Q_1,Q_2$ be $\alpha_1,\alpha_2,\alpha_3;\beta_1,\beta_2,\beta_3$. It's easily provable that it's impossible both polynomials to have $3$ real roots each. WLOG let $Q_1$ have $2$ complex (non-real) roots. We have that there $\exists$ a function $f:\alpha_{\{0,1,2\}}\rightarrow\beta_{\{0,1,2\}}$ s.t. $f(\alpha_i)-\alpha_i$ is an integer. We will prove that this function is injective from where it follows that it's bijective. Let FTSOC $f(\alpha_1)=f(\alpha_2)=\beta_1$. Then this means that $a_1-a_2\in\mathbb{Z}$. As one of the $2$ is complex (non-real) then the other one is it's complex conjugate, but 2 complex conjugates (non-real) don't have difference an integer. Now that we have that $f$ is a bijection, this means we can WLOG set $\beta_i=\alpha_i+n_i$, where $n_i\in\mathbb{Z}$. By Vieta's formulas we get $\alpha_1+\alpha_2+\alpha_3=-a$ and $\alpha_1+\alpha_2+\alpha_3+n_1+n_2+n_3=a \Rightarrow a\in\mathbb{Q}$. But from here we get by subtracting the following 2: $$\alpha_1^3+a\alpha_1^2+1=0$$$$ (\alpha_1+n_1)^3-a(\alpha_1+n_1)^2+1=0$$that $\alpha_1$ is a root of a quadratic equation. Getting rid of the $\alpha=-2$ we get that $Q_1$ is irreducible over $\mathbb{Q}$ from which our desired contradiction follows.