Finding the value of $3(\alpha-\beta)^2$ if $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all $f$

Let $f$ be a polynomial of degree $n$ at most $3$ such that there exists some $\alpha,\beta$ satisfying $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all such $f$. Find the value of $3(\alpha-\beta)^2$

Attempt: We need to find the value of $3(\alpha-\beta)^2$. Now, this expression reminds of the function $f(x) = (x-\beta)^3$. Differentiating this function at $x=\alpha$ will give us the desired result.

Hence, we just need to know the derivative of $k(x)=(x-\beta)^3$ at every point.

However, I am unable to move forward with this intuition. Could anyone please help me to move ahead in this problem.

Thank you very much for your help in this regard.

Solutions Collecting From Web of "Finding the value of $3(\alpha-\beta)^2$ if $\int_0^2 f(x)dx=f(\alpha) +f(\beta)$ for all $f$"

Since

$$\frac{8}{3} = \int_0^2 x^2dx = \alpha^2+\beta^2$$
$$2 = \int_0^2 x \ dx = \alpha + \beta$$

You get

$$(\alpha-\beta)^2 = 2(\alpha^2+\beta^2) -(\alpha+\beta)^2 = 2 \cdot \frac{8}{3} – 4 = \frac{4}{3}$$
so $3(\alpha-\beta)^2 = 4$

However, note that from this one is able to compute $\{\alpha, \beta\} = \{ 1\pm\frac{1}{\sqrt{3}}\}$. And now
$$\int_0^2 x^3 \ dx = 4 \neq \alpha^3 + \beta ^3$$

so, the formula can be valid only for polynomials of degree $\leq 2$.

$$f(x)=ax^3+bx^2+cx+d\\
\int_0^2f(x)dx=4a+8b/3+2c+2d\\
=a\alpha^3+b\alpha^2+c\alpha+d+a\beta^3+b\beta^2+c\beta+d\text{ for all a,b,c,d}\\
4=\alpha^3+\beta^3,8/3=\alpha^2+\beta^2,2=\alpha+\beta,2=1+1$$
Try to calculate $3(\alpha-\beta)^2$ from those equations.