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.
 A: We have $P(\alpha)=P(\beta)=0$, so we can express $b$ in terms of the distinct roots $\alpha$ and $\beta$
$$b=\frac{(1-\alpha^4)}{\alpha^3}=\frac{(1-\beta^4)}{\beta^3}$$
from which we can express $b^2$ as follows:-
$$b^2=\frac{(1-\alpha^4)(1-\beta^4)}{\alpha^3\beta^3} \text{ Eq.(1)}$$
In addition we obtain the following relationship, which can be further manipulated 
$$\beta^3(1-\alpha^4)=\alpha^3(1-\beta^4)\\\Rightarrow \alpha^3\beta^3=\frac{\alpha^3-\beta^3}{\beta-\alpha}=-(\alpha^2+\alpha\beta+\beta^2)\\\Rightarrow(\alpha^3\beta^3+\alpha\beta)^2=(-(\alpha^2+\beta^2))^2\\\Rightarrow \alpha^6\beta^6+2\alpha^4\beta^4+\alpha^2\beta^2=\alpha^4+2\alpha^2\beta^2+\beta^4\\\Rightarrow \alpha^6\beta^6+2\alpha^4\beta^4-\alpha^2\beta^2-(\alpha^4+\beta^4)=0 \text{ Eq.(2)}$$
Using Equation (1) then (2), we can express $Q(\alpha\beta)$ as follows:-
$$Q(\alpha\beta)=\alpha^6\beta^6+\alpha^4\beta^4+\color{red}{\frac{(1-\alpha^4)(1-\beta^4)}{\alpha^3\beta^3}}\alpha^3\beta^3-\alpha^2\beta^2-1\\=\alpha^6\beta^6+2\alpha^4\beta^4-\alpha^2\beta^2-(\alpha^4+\beta^4)=0$$
A: $P(\alpha)=0$ 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
