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?

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 $i<j$.
