Let $P(x)$ be any polynomial of degree at most $3$.$\int_{-1}^{1}P(x)dx = P(x_1)+P(x_2)$, where $x_1$ and $x_2$ are independent of the polynomial P. Let $P(x)$ be any polynomial of degree at most $3$. It can be shown that there
are numbers $x_1$ and $x_2$ such that $$\int_{-1}^{1}P(x)\,\operatorname{d}x = P(x_1)+P(x_2),$$ where $x_1$ and $x_2$ are independent of the polynomial $P$.


*

*Show that $x_1=−x_2$.

*Find $x_1$ and $x_2$.



I let $P(x) := ax^3+bx^2+cx+d$
According to the question,
$$\int_{-1}^{1}\left(ax^3+bx^2+cx+d\right)\,\operatorname{d}x=ax_1^3+bx_1^2+cx_1+d+ax_2^3+bx_2^2+cx_2+d$$
$$\frac{2b}{3}+2d=ax_1^3+bx_1^2+cx_1+d+ax_2^3+bx_2^2+cx_2+d$$
I am stuck here. I dont know how to solve further. Please help me.

You already have two fine answers, under the assumption that such $x_1, x_2$ exist.  Your own argument can be completed to show that they do indeed exist.  You have already reached the equivalent statement:
$$a(x_1^3+x_2^3) + b(x_1^2+x_2^2-\frac23) + c(x_1 + x_2)=0$$
which must hold for all $a, b, c$.  Clearly $x_1+x_2=0$ will ensure it holds for any $a, c$, and then you need $x_1^2+x_2^2=\frac23$ if it has to hold for any $b$.  
Solving these two conditions, we do get two real solutions - i.e $x_1 = -x_2 = \pm \frac1{\sqrt3}$, so in fact we have also shown the existence.
