<s>[INST] If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$. If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$.
Students and I are unsure how to go about this problem. Also will this be a problem I can solve and prove in front of a class in 20 minutes?
 [/INST] There is probably a shorter way, but I think this way sheds some light on why it's true:
Let $a,b,c,d$ be the four roots of $x^4+x^3-1$, so that $x^4+x^3-1=(x-a)(x-b)(x-c)(x-d)$. Now set
$$
g(x) = (x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd).
$$
This polynomial is symmetric in the roots $a,b,c,d$, and so its coefficients will be rational numbers (since the coefficients of $x^4+x^3-1$ are rational). This explains why there is such a sextic; to work out its coefficients, we need to mess around with symmetric polynomials.
The coefficients of the original polynomial tell us that $a+b+c+d=-1$, $abcd=-1$, and $ab+ac+ad+bc+bd+cd=abc+abd+acd+bcd=0$. Therefore the coefficient of $x^5$ in $g(x)$ equals $0$; the coefficient of $x^4$ equals
$$
abac+abad+abbc+abbd+abcd+acad+acbc+acbd+accd+adbc+adbd+adcd+bcbd+bccd+bdcd = (a + b + c + d) (a b c + a b d + a c d + b c d) - abcd = 1;
$$
and so on - the coefficient of $x^0$ equals $(abcd)^3=-1$.
(One can play the same game with $f(x)=(x-a^2)(x-b^2)(x-c^2)(x-d^2)$ to obtain the polynomial $x^4−x^3−2x^2+1$.)
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