If $\int_{0}^{2} p(x) dx = p(\alpha) + p(\beta)$ for all polynomials of degree at most $3$, what's $3(\alpha - \beta)^2$? If for some $\alpha , \beta \in \mathbb{R}$ , the integration formula 
$\int_{0}^{2} p(x) dx = p(\alpha) + p(\beta)$ , holds for all polynomials $p(x)$ of degree atmost $3$, then the valueof $3(\alpha - \beta)^2 = ?$.
How to approach this question? i thought of taking Polynomial as $x^3$ and i am getting the answer but is there any formal way of doing it without explicitly substituting any polynomial of degree atmost $3$ ?.
Any suggestion , hints , solution ?
 A: Following Isko10986's suggestion, if we take $p(x)=p_3x^3+p_2x^2+p_1x+p_0$, we have 
$$\int_0^2 p(x)dx =\frac{2^4}{4}p_3+\frac{2^3}{3}p_2+\frac{2^2}{2}p_1+2p_0 \\ =4p_3+\frac{8}{3}p_2+2p_1+2p_0$$
Given that it is equal to $p(\alpha)+p(\beta)=(p_3\alpha^3+p_2\alpha^2+p_1\alpha+p_0)+(p_3\beta^3+p_2\beta^2+p_1\beta+p_0)$ 
we conclude that $$4p_3+\frac{8}{3}p_2+2p_1+2p_0=(p_3\alpha^3+p_2\alpha^2+p_1\alpha+p_0)+(p_3\beta^3+p_2\beta^2+p_1\beta+p_0)$$
is an identity in $p_3,p_2,p_1,p_0$.
So we have got that,
$$\alpha^3+\beta^3=4\tag1$$
$$\alpha^2+\beta^2=\frac{8}{3}\tag2$$
$$\alpha+\beta=2\tag3$$
From $(3)$ and $(2)$ , we get that $\alpha\beta=\frac{2}{3}$ and hence $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\frac{4}{3}$
So we have $$\alpha=1+\frac{1}{\sqrt3}$$ and $$\beta=1-\frac{1}{\sqrt3}$$
So as you can see, this is the most formal way possible for we have assumed a generalised polynomial of degree $3$.
And $\boxed{3(\alpha-\beta)^2=4}$

Now, for the second part of your question, yes there does not exist any $\alpha,\beta \in \mathbb{R}$ such that 

$\int_{0}^{2} p(x) dx = p(\alpha) + p(\beta)  $ , holds for all polynomials $p(x)$ of degree atmost $n > 3$

I can provide a glimpse of this from the above procedure again.
Suppose we take $p(x)=\sum_\limits{k=0}^n p_kx^k$
So $$\int_0^2 p(x)dx =\sum_\limits{k=0}^n p_k\frac{2^{k+1}}{k+1}$$
and the $n$ equations in this case will be given by $$\alpha^k+\beta^k=\frac{2^{k+1}}{k+1} \,\,\,\,\,\, k=1,2,3, \ldots, n$$
However, since there are only $2$ variables, solving the equations for $k=1,2$ will give the same above mentioned values of $\alpha$ and $\beta$ and the other equations will be redundant .
And here is the twist:
So here also, you get $$\alpha=1+\frac{1}{\sqrt3}$$ and $$\beta=1-\frac{1}{\sqrt3}$$
And the equation for $k=4$ gives $\alpha^4+\beta^4=\frac{32}{5}$ and the above solution does not satisfy this.
So this representation is possible only for $n \le 3, n\in \mathbb{N}$.
Hope this helps you.
A: A quick way to get the answer (assuming that there do indeed exist constants such that the formula holds, which is what you set out to formally prove) is the following:
Note that $(\alpha-\beta)^2+(\alpha+\beta)^2 = 2(\alpha^2+\beta^2)$, and hence $3(\alpha-\beta)^2 = 6s_2 - 3s_1^2$, where $s_1 = \alpha+\beta$ and $s_2 = \alpha^2+\beta^2$. Notice, however, that $s_1 = \int\limits_{0}^{2}{x\,dx}$ and $s_2 = \int\limits_{0}^{2}{x^2\,dx}$.
A: Solving the integral, $\int_0^2 (a_3x^3+a_2x^2+a_1x+a_0)dx = 4a_3+\frac{8}{3}a_2+2a_1+2a_0$
Meanwhile, 
$p(\alpha)+p(\beta)=a_3(\alpha^3+\beta^3)+a_2(\alpha^2+\beta^2)+a_1(\alpha+\beta)+2a_0$
Equate; compare coefficients; solve for what you want.
