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


Lemma: If $\alpha,\beta,\gamma$ are the roots of an irreducible polynomial over $\mathbb{Q}$ with degree $3$, $\alpha\beta$ is a root of: 
  $$ q(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma)\in\mathbb{Q}[x]. $$

Proof: The coefficients of $q(x)$ are symmetric polynomials in $\alpha,\beta,\gamma$, hence $q(x)\in\mathbb{Q}[x]$.
The same argument also shows that:

Lemma: If $\alpha,\beta,\gamma,\delta$ are the roots of an irreducible polynomial over $\mathbb{Q}$ with degree $4$, $\alpha\beta$ is a root of: 
  $$ q(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\delta)(x-\beta\delta)(x-\gamma\delta)\in\mathbb{Q}[x]. $$

By applying Viète's formulas to the polynomial $x^4+x^3-1$, we have:
$$\alpha+\beta+\gamma+\delta = -1, $$
$$\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=0, $$
$$\alpha\beta\gamma\delta = -1,$$
$$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}=\frac{1}{\alpha\beta}+\frac{1}{\alpha\gamma}+\frac{1}{\alpha\delta}+\frac{1}{\beta\gamma}+\frac{1}{\beta\delta}+\frac{1}{\gamma\delta}=0,$$
so the elementary symmetric polynomials of $\alpha\beta,\alpha\gamma,\alpha\delta,\beta\gamma,\beta\delta,\gamma\delta$ are straightforward to compute, giving $q(x)=x^6+x^4+x^3-x^2-1$ as wanted.
