$Q(x)$ be a polynomial of degree $n$.
$$
Q(1)=1 \text { and } \frac{Q(2 x)}{Q(x+1)}+\frac{56}{x+7}-8=0...(i)
$$
Put $x=0$ in Eq. (i), we get
$$
\begin{aligned}
\frac{Q(0)}{Q(1)}+\frac{56}{7}-8 & =0 \\
\Rightarrow \quad \frac{Q(0)}{1}+8-8 & =0 \\
Q(0) & =0
\end{aligned}
$$
Solving Eq. (i), we get
$$
\begin{aligned}
& \frac{Q(2 x)}{Q(x+1)}+\frac{56-8 x-56}{x+7}=0 \\
& \Rightarrow \quad \frac{Q(2 x)}{Q(x+1)}-\frac{8 x}{x+7}=0 \\
& \Rightarrow \quad \frac{Q(2 x)}{Q(x+1)}=\frac{8 x}{x+7}...(ii) \\
& \because \quad Q(0)=0 \\
&
\end{aligned}
$$
$\therefore \quad x$ is a factor of $Q(x)$.
Let $Q(x)=x P(x)$
$$
\therefore \quad \begin{aligned}
\frac{Q(2 x)}{Q(x+1)} & =\frac{8 x}{x+7} \\
\frac{2 x \cdot P(2 x)}{(x+1) P(x+1)} & =\frac{8 x}{x+7} \\
\frac{P(2 x)}{P(x+1} & =\frac{4(x+1)}{x+7}...(ii)
\end{aligned}
$$
Put $x=-1$
$$
\frac{P(-2)}{P(0)}=0 \Rightarrow P(-2)=0
$$
$\Rightarrow \quad x+2$ is factor of $P(x)$
$$
P(x)=(x+2) R(x)
$$
Now, in Eq, (ii), we get
$$
\begin{aligned}
\frac{4(x+1)}{x+7} & =\frac{P(2 x)}{P(x+1)}=\frac{(2 x+2) R(2 x)}{(x+1+2) R(x+1)} \\
\Rightarrow \quad \frac{4(x+1)}{x+7} & =\frac{2 x+2}{x+3} \frac{R(2 x)}{R(x+1)} \\
\frac{2(x+3)}{x+7} & =\frac{R(2 x)}{R(x+1)} \\
\because \quad x & =-3 \Rightarrow 0=\frac{R(-6)}{R(-2)} \Rightarrow R(-6)=0
\end{aligned}
$$
$=x+6$ is a factor of $R(x)$.
So, $R(x)=(x+6) S(x)$
$$
\begin{aligned}
\frac{2(x+3)}{x+7} & =\frac{R(2 x)}{R(x+1)}=\frac{(2 x+6) S(2 x)}{(x+6+1) S(x+1)} \\
\Rightarrow & \frac{S(2 x)}{S(x+1)}=1 \Rightarrow S(2 x)=S(x+1) \text { for all } x
\end{aligned}
$$
$\Rightarrow S(x)$ is constant function.
$$
\therefore Q(x)=x P(x)=x(x+2) R(x)=x(x+2)(x+3) S(x)
$$
$\therefore Q(x)$ is cubic polynomial.
$$
\begin{aligned}
& \therefore n=3 \\
& \begin{aligned}
\text { Required value } & ={ }^3 C_0+{ }^3 C_1+{ }^3 C_2+{ }^3 C_3 \\
& =1+3+3+1=8
\end{aligned}
\end{aligned}
$$