Let $a, b, c$ be non-zero real numbers, such that $\displaystyle\int_{0}^{1}(1 + \cos^8 x)(ax^2 + bx + c)\,dx = \displaystyle\int_{0}^{2}(1 + \cos^8 x)(ax^2 + bx + c)\,dx$, then $ax^2 + bx + c = 0$ has

Given $\displaystyle\int_{0}^{1}(1 + \cos^8 x)(ax^2 + bx + c)\,dx = \displaystyle\int_{0}^{2}(1 + \cos^8 x)(ax^2 + bx + c)\,dx$

$\Rightarrow \displaystyle\int_{1}^{2}(1 + \cos^8 x)(ax^2 + bx + c)\,dx = 0$

Let $f(x) = 1 + \cos^8 x$ and $g(x) = ax^2 + bx + c$.

We know that $f(x) \ge 1 > 0$ for all $x \in \mathbb{R}$.

If $g(x)$ does not change sign in the interval $(1, 2)$, then the product $f(x)g(x)$ will also not change sign in $(1, 2)$. Since $g(x)$ is a non-zero polynomial, the definite integral of $f(x)g(x)$ over $(1, 2)$ would be strictly positive or strictly negative, which contradicts the fact that the integral is zero.

Therefore, $g(x)$ must change sign in the interval $(1, 2)$.

Since $g(x)$ is a continuous function, by the Intermediate Value Theorem, $g(x) = 0$ must have at least one root in $(1, 2)$.

Answer: at least one root in $(1, 2)$