So, I tried vieta and with some algebra, I got to the point $0=\alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \gamma+\gamma \beta+\alpha \beta)$ where $(\alpha \gamma+\gamma \beta+\alpha \beta)=a$. I don't know how to progress from here
$f(x)=x^{3}-3 x+a$. Given that it has $3$ integer roots $\alpha, \beta, \gamma$. Find all possible values of $a$.
algebra-precalculus
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Best Answer
We can solve this without breaking symmetry, $$\alpha+\beta+\gamma=0$$ squaring this we have $$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma)=0$$ So $$\alpha^2+\beta^2+\gamma^2=6$$ now since $\alpha,\beta, \gamma$ are integers we have the only possible values of $\alpha^2,\beta^2,\gamma^2$ are $0, 1, 4$. And it is easy to see that we must have, up to a permutation,
$$\alpha^2=\beta^2=1, \gamma^2=4$$ and thus
$$\alpha^2\beta^2\gamma^2=4$$ so
$$a=\sqrt{4}=\pm 2$$