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

Let $a,b,c,d$ be the roots of $x^4+x^3-1=0$.
By Vieta's formula,
$$a+b+c+d=-1\quad\Rightarrow\quad c+d=-1-(a+b)\tag1$$
$$abcd=-1\quad\Rightarrow \quad cd=-\frac{1}{ab}\tag2$$
Since we have
$$a^4+a^3=1\quad\text{and}\quad b^4+b^3=1$$
we can have
$$1=(a^4+a^3)(b^4+b^3)$$
$$(ab)^4+(ab)^3(a+b+1)=1,$$
i.e.
$$a+b=\frac{1-(ab)^4}{(ab)^3}-1\tag3$$
Similarly, 
$$(cd)^4+(cd)^3(c+d+1)=1$$
Using $(1)(2)$, this can be written as
$$\left(-\frac{1}{ab}\right)^4+\left(-\frac{1}{ab}\right)^3(-1-(a+b)+1)=1,$$
i.e.
$$a+b=\left(1-\frac{1}{(ab)^4}\right)(ab)^3\tag4$$
From $(3)(4)$, letting $ab=x$, we have
$$\frac{1-x^4}{x^3}-1=\left(1-\frac{1}{x^4}\right)x^3$$
to get
$$x^6+x^4+x^3-x^2-1=0.$$
