If $a$ and $b$ are roots of $x^4+x^3-1=0$, $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I have to prove that:

If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.

I tried this :
$a$ and $b$ are root of $x^4+x^3-1=0$ means :
$\begin{cases}
a^4+a^3-1=0\\
b^4+b^3-1=0
\end{cases}$
which gives us :
$(ab)^4+(ab)^3=a^3+b^3+a^4+b^4+a^4b^3-a^3b^4-1$
can you help me carry on ? or propose another solution ? thanks in advance

The companion matrix of $p(z) = z^4 + z^3 - 1$ is
$$ A = \pmatrix{0 & 0 & 0 & 1\cr
                1 & 0 & 0 & 0\cr
                0 & 1 & 0 & 0\cr
                0 & 0 & 1 & -1\cr}$$
This is a matrix whose eigenvalues are the roots of $p(z)$.  The Kronecker product $A \otimes A$ is a $16 \times 16$ matrix; all products $ab$ where $a$ and $b$ are roots of $p(z)$ are eigenvalues of $A \otimes A$.  Moreover, if $a \ne b$, $ab$ will be an eigenvalue of multiplicity $\ge 2$, because 
if $u$ and $v$ are eigenvectors of $A$ for eigenvalues $a$ and $b$, 
both $u \otimes v$ and $v \otimes u$ are eigenvectors of $A \otimes A$ for eigenvalue $ab$.  
Now the 
characteristic polynomial of $A \otimes A$ turns out to be
$$\left( {z}^{4}-{z}^{3}-2\,{z}^{2}+1 \right)  \left( {z}^{6}+{z}^{4}+{
z}^{3}-{z}^{2}-1 \right) ^{2}
$$
The products $ab$ with $a \ne b$ are roots of this characteristic polynomial
with multiplicity $\ge 2$, and therefore are roots of $ {z}^{6}+{z}^{4}+{
z}^{3}-{z}^{2}-1 $.  
In the case $a=b$, $ab = a^2$ turns out to be a root of the other factor
$ {z}^{4}-{z}^{3}-2\,{z}^{2}+1 $.
