Let $\mathbb F_q$ be a finite field (with $q$ not a power of $2$ or $3$). Let $\alpha$ be an element of $\mathbb F_{q^3}$ having irreducible minimal polynomial of degree $3$ over $\mathbb F_q$. Can you give a closed formula in terms of the coefficients of the minimal polynomial of $\alpha$ over $\mathbb F_q$ for $Tr_{\mathbb F_{q^3}/\mathbb F_q}(\alpha^{q+2}) $?
The coefficients of the minimal polynomial of alpha are clearly $Tr(\alpha)$, $N(\alpha)$ and $Tr(\alpha^{q+1})$ but I am unsure how to combine those to get the wanted formula.
I suspect not. Let $\beta=\alpha^q$ and $\gamma=\beta^q$ (so $\alpha=\gamma^q$). Then (I'll use $T$ for trace) $$\tau=T(\alpha^{q+2})=T(\alpha^2\beta)=\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha.$$ Alas this is not a symmetric polynomial in $\alpha$, $\beta$, $\gamma$. But if we let $$\tau'=\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2$$ then $(\tau-\tau')^2$ is the discriminant of $f$. Also $$\tau+\tau'=(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) -3\alpha\beta\gamma=T(\alpha)T(\alpha^{q+1})N(\alpha).$$ So we can write down a quadratic with coefficients given in terms of $T(\alpha)$, $T(\alpha^{q+1})$ and $N(\alpha)$ having $\tau$ as a root. But I cannot see a simple way of telling which root is $\tau$.