5. Find all positive real numbers $a, b, c, d$ such that for all $x \in \mathbb{R}$,
$$
(a x+b)^{2016}+\left(x^{2}+c x+d\right)^{1008}=8(x-2)^{2016} .
$$
Answer: $a=7 \frac{1}{2016}, b=-2 \cdot 7 \frac{1}{2016}, c=-4, d=4$
Solution: Compare coefficients of $x^{2016}$ in the equation to obtain $a^{2016}+1=8$, i.e. $a=7 \frac{1}{2016}$. Then, take $x=2$ to obtain
$$
(2 a+b)^{2016}+(4+2 c+d)^{1008}=0 .
$$

Since the LHS is a sum of even-exponent powers, the equation will be solved in $\mathbb{R}$ if and only if both addends are zero. In particular, $b=-2 a=-2 \cdot 7 \frac{1}{2016}$. Finally, substitute these values for $a$ and $b$ in the original equation to obtain
$$
\begin{aligned}
7(x-2)^{2016}+\left(x^{2}+c x+d\right)^{1008} & =8(x-2)^{2016} \\
\left(x^{2}+c x+d\right)^{1008} & =\left(x^{2}-4 x+4\right)^{1008}
\end{aligned}
$$

Comparing coefficients once more, we finally obtain $c=-4$ and $d=4$.