<s>[INST] Prove that, for every $a\neq0$, the equations $ax^3 - x^2 -x -(a +1)= 0$ and $ax^2 - x -(a +1)=0$ have a common root 
Show that, for every real number $a\neq0$, the equations
$ax^3 - x^2 -x -(a +1)= 0$ and $ax^2 - x -(a +1)=0$
have a common root.

I want to know whether there are any mistakes in it.
Proof: If the system of equations
$$ax^3-x^2-x-(a+1)=0\cdots(1)$$ and $$ax^3-x^2-x=ax^2-x\cdots(2)$$ is considered,let us see what roots are obtained. Now,Subtracting (1) and (2) we get,
$$-(a+1)=x-ax^2\Rightarrow ax^2-x-(a+1)=0$$ This equation has roots $-1$ and $(1+2a+1)\times \frac{1}{2a}=\frac {a+1}{a}$.We see that $\frac {a+1}{a}$ is the root of (1) and (2) but (-1) is not(we can verify that). Thus, we found a common root for $a$ for all $a\in \mathbb R$ with $a\neq0$.
 [/INST] You have almost finished the solution.  But you're missing one step.
Because, we must check that the solution satisfies at least one equation.
You can also reach the result as follows:
$$\begin{align}&\begin{cases}ax^3-x^2-x-(a+1)=0\\ ax^2-x-(a+1)=0\end{cases} \\\\ \implies &\begin{cases}ax^3-x^2-x-(a+1)=0\\ ax^3-x^2-x(a+1)=0\end{cases}\\
\implies &x(a+1-1)=a+1\\
\implies &x=\frac{a+1}{a} \end{align}$$
Finally, to conclude the proof, it is sufficient to check whether the root we found satisfies the quadratic equation:
$$\frac{(a+1)^2}{a}-\frac{a+1}{a}-(a+1)=0.$$
This means, you're done.
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