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

Question

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 ?

Answer 1

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 $$ 64a_1+27a_2+8a_3+a_4=729. $$

Answer 2

As @Henno Brandsma mentioned in the comment, comparing the coefficients of $x^3$ in $(1)$ gives you $a_1+a_2+a_3+a_4=8$. The second equation can be obtained by setting $x=-5$ in equation $(1)$.

Answer 3

The LHS and the RHS of $(1)$ are two third-degree polynomials that agree over $0,1,2,3$ due to the given constraints. It follows that they are the same polynomial, hence they have the same leading term and the same value at $-5$.

Answer 4

The blunt-force approach:

The system you mention is simply $$\left(\begin{array}{cccc} 1 & 8 & 27 & 64\\ 8 & 27 & 64 & 125\\ 27 & 64 & 125 & 216\\ 64 & 125 & 216 & 343\\ \end{array}\right)\left(\begin{array}{c}a_1\\a_2\\a_3\\a_4\end{array}\right)=\left(\begin{array}{c}1\\27\\125\\343\end{array}\right)$$ or $Aa=b$, with solution $a=A^{-1}b=\left(\begin{array}{c}17.5\\-17.5\\10.5\\-2.5\end{array}\right).$ So indeed $a_1+a_2+a_3+a_4=8$ and $64a_1+27a_2+8a_3+a_4=729$.

Then simply writing out both sides should give you equality in equation (1).