Simplifying $\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)}$ using Lagrange’s polynomial

Question

I have a question of symplifying this expression $$ A=\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \tag{1} $$ where $a$, $b$ and $c$ are distinct nonzero real numbers. There is a quite similar problem, which is simplifying $$ B=\frac{a^2}{(a-b)(a-c)} + \frac{b^2}{(b-a)(b-c)} + \frac{c^2}{(c-a)(c-b)}. \tag{2} $$ In this problem, they use Lagrange’s interpolation polynomial for $P(x) =x^2$ at nodes $a$, $b$ and $c$. Then, they compare the coefficient of $x^2$ to obtain that $B=1$.

My effort is to use the similar method with $P(x)=x^4$, but I am struggling with the fact that there are only three nodes here. So the polynomial I need to choose must be a second degree polynomial but it is hard for me to find such one.

Can someone have some idea about using different polynomial to use the Lagrange’s interpolation polynomial? Thank you very much!

Answer 1

You want to write down a quadratic polynomial $p$ for which $p(a)=a^4$, $p(b)=b^4$, and $p(c)=c^4$.

We see that $p(x)=x^4$ whenever $x\in\{a,b,c\}$. This means that $$(x-a)(x-b)(x-c)\mid (x^4-p(x)).$$ In particular, we can reduce $x^4$ modulo $(x-a)(x-b)(x-c)$, and we'll have the polynomial we need.

Answer 2

For your problem, I threw it at Wolfram Alpha and got

$a^2 + a b + a c + b^2 + b c + c^2 $.

This can be rewritten as $(a+b+c)^2-(ab+ac+bc)$.

With 3 instead of 4 this is $a+b+c$, and with 5 this is $a^3 + a^2 b + a^2 c + a b^2 + a b c + a c^2 + b^3 + b^2 c + b c^2 + c^3 $.