<s>[INST] show that $a_1+a_2+a_3+a_4=8$ and that $64a_1+27a_2+8a_3+a_4=729$ given the following 
Consider the sistem of equations:  $$\begin{cases} 
 a_1+8a_2+27a_3+64a_4=1 \\ 8a_1+27a_2+64a_3+125a_4=27 \\
 27a_1+64a_2+125a_3+216a_4=125\\ 64a_1+125a_2+216a_3+343a_4=343\\
 \end{cases} $$ These four equations determine $a_1,a_2,a_3,$and $a_4$.
Show that $$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3 \equiv
 (2x+1)^3,\tag 1$$i.e.,these two polynomials are identically the same.
Use this to show that $a_1+a_2+a_3+a_4=8$ and that
  $64a_1+27a_2+8a_3+a_4=729$.

I was able to prove the first part but I am having trouble  showing that $a_1+a_2+a_3+a_4=8$.
My approach was to calculate equation $(1)$ when $x=0,-1,-2,-3$ ,thus yielding:
$a_1=1-(8a_2+27a_3+64a_4) $
$a_2=-1-(8a_3+27a_4)$
$a_3=-27-(-a_1-8a_4)$
$a_4=-125-(-8a_1-a_2)$
So when I add them I get $$a_1+a_2+a_3+a_4=-27-125+(9a_1-7a_2-35a_3-91a_4)$$
The problem is that I don't see how to simplify  $9a_1-7a_2-35a_3-91a_4$ to get that it equals $160$.
Can you guys give me some hint ?
 [/INST] Consider $p(x)=a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3 -
 (2x+1)^3$ then $p(0)=p(1)=p(2)=p(3)=0$ that is the polynomial $p(x)$ of degree at most 3 have 4 distinct zeros. Thus $p(x)\equiv0$ i.e., $$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3 \equiv
 (2x+1)^3.~~~~~~~~(1)$$
Comparing coefficients of $x^3$ on b/s of $(1)$ we get
$$ a_1+a_2+a_3+a_4=8 $$
Put $x=-5$ in $(1),$ we get
$$ 64a_1+27a_2+8a_3+a_4=729. $$
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