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I am asked the following question:

Consider a field $K$ with characteristic different from 2 and 3, and the polynomial $f(t) = t^3 + pt + q \in K(t)$ with three distinct roots $\alpha_1, \alpha_2, \alpha_3 \in \overline{K}$ ($\overline{K}$ denotes the algebraic closure of $K$). Show that $$ \alpha_1 + \alpha_2 + \alpha_3 = 0 $$

Then there are similar questions about the squares and the cubes of the roots, but I am already stuck with this one. I tried using the fact that $\alpha_i^3 + p\alpha_i + q=0$ and manipulating this expression, but I did not get anywhere.

Could you give me some hints?

Thanks in advance,

Olivier

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Hint If $\alpha_1, \alpha_2, \alpha_3$ are the roots of $f(t)$, then $f(t) = (t-\alpha_1)(t-\alpha_2)(t-\alpha_3)$. Multiply this out and compare to the definition of $f(t)$ with which you started.

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