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?
 A: The resultant
$$h(z)=\text{Res}( \ \text{Res} ( \ z-xy, \ x^4+x^3-1, \ x), \ y^4+y^3-1, \ y)$$ 
is a $16^{th}$degree  polynomial in $z$ with roots $\{a_ia_j:i,j=1,2,3,4\}$, where $a_1,a_2,a_3,a_4$ are the roots of $x^4+x^3-1$.
By expanding the above resultant we get that the polynomial $h(z)$ is $$h(z)= \left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right)  \left( {z}^{6}+{z}^{4}+{z}^{3}-{z}^{2}-1 \right)^2.$$
The roots of the polynomial $\left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right)$ are $a_i^2$ and the roots of $\left( {z}^{6}+{z}^{4}+{z}^{3}-{z}^{2}-1 \right)$ are $a_ia_j$ with $i<j$.
A: The statement holds true only for distinct roots $\rm\, a\ne b.\:$ As such, the proof will require use of $\rm\, a\ne b,\:$ e.g. by cancelling $\rm\,a-b.\:$ Let $\rm\,f(x)\,$ be the quartic and let $\rm\,g(x)\,$ be the sectic. Then $\rm\,g(ab)\,$ can be reduced to $\,0\,$ by using $\rm\,\color{#C00}{f(a)} = 0 = \color{#C00}{f(b)}\ $ and $\rm\ \color{#C00}{h(a,b)} = \dfrac{f(a)-f(b)}{a-b} =\, 0,\,$ namely
$$\rm\begin{eqnarray} g(ab)\,\ &=&\ \rm (a^5\!+ a^4b+ a^3b^2\!+\!a^2 b^2\!+a^3\!+1)\ \color{#C00}{f(a)} \\
&&  \ +  \rm\  (a^6 b^2\!-\!a^6 b+a^6\!+a^4)\ \color{#C00}{f(b)}\\
 &&\ -\  \rm (a^6\!+a^4\!-\!a^3)\ \color{#C00}{ h(a,b)}\\
 &=&\ \ 0\quad\rm 
\end{eqnarray}$$
Remark $\ $ There may well be a simpler such expression (e.g. a symmetric one). I have not attempted to simplify it.
A: 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 $x^4−x^3−2x^2+1$.)
