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?

Solution 1:

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 $x^4−x^3−2x^2+1$.)

Solution 2:

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 $i<j$.