<s>[INST] Find the value of $Q(x)$ at $x= -1$, knowing some of its properties. 
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
  $$
\frac{Q(2x)}{Q(x+1)}= 8 -\frac{56}{x+7},
\quad \forall x \ne -7 \text{ and } Q(x+1)\neq 0\,.
$$
  Find $Q(-1)$.

 [/INST] Rewrite the equation as
$$(x+7)\,Q(2x)=8x\,Q(x+1)\,.\tag{*}$$
Thus, $2x\mid Q(2x)$ or $x\mid Q(x)$, and $x+7\mid Q(x+1)$ or $x+6\mid Q(x)$.  That is, $Q(x)=x(x+6)\,R(x)$ for some polynomial $R(x)$.  Show that
$$(x+3)\,R(2x)=2(x+1)\,R(x+1)\,.$$
Ergo, $2x+2\mid R(2x)$ or $x+2\mid R(x)$, and $x+3\mid R(x+1)$ or $x+2\mid R(x)$.  Thence, $R(x)=(x+2)\,S(x)$ for some polynomial $S(x)$.  Then, prove that
$$S(2x)=S(x+1)\,,$$
which implies that $S(x)$ is a constant polynomial.  Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.

You can also notice that $\lim\limits_{x\to\infty}\,\dfrac{Q(2x)}{Q(x+1)}=8=\left(\dfrac{2}{1}\right)^3$, so $Q(x)$ must be cubic.  If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out.  There are some easily observable results.  Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$.  With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0\,,$$ and so on.
</s>