<s>[INST] Showing that $\alpha \beta$ is the root of a polynomial Assuming that $\alpha, \beta$ are distinct roots of $P(x) = x^4+bx^3-1 = 0$, where $b \in \mathbb R$, show that $\alpha \beta$ is the root of $Q(x) = x^6+x^4+b^2x^3 -x^2 -1$.
I have already noticed that $P(0) < 0$ and $\deg P = 4$, so (from the intermediate value theorem) there are at least 2 real roots. It seems that the third and fourth root are complex, but I'm not sure, if this could help.
 [/INST] $P(\alpha)=0$ and $P(\beta)=0$ therefore $\alpha^4=1-b\alpha^3$ and $\beta^4=1-b\beta^3$; multiplying by $\alpha^2$ (respectively$\beta^2$) we get $\alpha^6=-b\alpha^3+\alpha^2-b\alpha+b$ (resp $\beta^6=-b\beta^3+\beta^2-b\beta+b$) substitute the sixth and the fourth powers of $\alpha$ and $\beta$ in the expression of $$Q(\alpha\beta)=\alpha^6\beta^6+\alpha^4\beta^4+b^2\alpha^3\beta^3-\alpha^2\beta^2-1$$ simplify and you should get zero
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