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$$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}\equiv{a+b+c}$$

Demonstrate the identities above considering a, b & c real numbers and distinct from each other.

I'm stucked on this problem. Take a look on what I did:

$x=(a-b)\therefore -x=(b-a)$; $y=(a-c);-y=(c-a)$; $z=(b-c);-z=(c-b)$.

___________________________________________________________________$$\frac{a^3}{xy}+\frac{b^3}{(-x)z}+\frac{c^3}{(-y)(-z)}$$

$$\frac{a^3z-b^3y+c^3x}{xyz}$$

$$\frac{a^3(b-c)+b^3(a-c)+c^3(a-b)}{xyz}$$

$$\frac{a^3(b-c)+b^3(\color{red}{-b}+a-c\color{red}{+b})+c^3(a-b)}{xyz}$$

$$\frac{a^3b+ab^3-a^3c+ac^3-b^3c-bc^3}{xyz}$$

$$\frac{ab(a^2+b^2)-ac(a^2-c^2)-bc(b^2+c^2)}{xyz}$$

after this step I was thinking about doing the same technique on the second term, $-ac(\color{red}{b^2}+a^2-c^2\color{red}{-b^2})$, but after doing some more algebraic factorization I got stucked, could you help me to demonstrate?

Nicolas Leskiu
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2 Answers2

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Put the LHS over a common denominator, getting

$$\frac{a^3(b-c)}{(a-c)(b-c)(a-b)}-\frac{b^3(a-c)}{(a-c)(b-c)(a-b)}+\frac{c^3(a-b)}{(a-c)(b-c)(a-b)}=$$

$$=\frac{a^3b-a^3c-ab^3+ac^3+b^3c-bc^3}{(a-c)(b-c)(a-b)}=\frac{(a+b+c)(a-c)(b-c)(a-b)}{(a-c)(b-c)(a-b)}=a+b+c$$

vadim123
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Assuming $b\neq c$ the function $$ f(z) = \frac{z^3}{(z-b)(z-c)}+\frac{b^3}{(b-c)(b-z)}+\frac{c^3}{(c-z)(c-b)} $$ is holomorphic over $\mathbb{C}\setminus\{b,c\}$. In the worst scenario $b$ and/or $c$ are simple poles, but $\operatorname*{Res}_{z=c}f(z)=\operatorname*{Res}_{z=b}f(z)=0$ imply that $f$ is entire (once its removable discontinuities are removed). Given that, by inspecting the definition above it is clear that $f(z)$ is a polynomial with degree $\leq 1$, and actually a linear polynomial of the form $f(z)=z+q$. Since $$ f(0) = \frac{b^2}{b-c}+\frac{c^2}{c-b} = \frac{b^2-c^2}{b-c} = b+c $$ we have $f(z)=z+b+c$ for any $z\in\mathbb{C}\setminus\{b,c\}$, in particular for $z=a$.

Jack D'Aurizio
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  • Dude, I just need to simplify it till gets to a+b+c. Not this math that I don't understand. – Nicolas Leskiu Mar 21 '18 at 22:08
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    This math you don't understand allows you to prove such statement without performing heavy computations, just relying on general principles. I recall that answers on MSE also serve as a repository of knowledge, they are not mere homework-solving tools. In my opinion, you should be intrigued by meeting new math, not just stick to the given homework. – Jack D'Aurizio Mar 21 '18 at 22:13
  • Man, my objective is to crush those kind of problems, then I will have the ability to understand more complicated topics. Can't you see that? – Nicolas Leskiu Mar 21 '18 at 22:17
  • I'll be glad if you gimme a hint on how to simplify it by hard work – Nicolas Leskiu Mar 21 '18 at 22:18
  • @NicolasLeskiu: in a elementary way: adjust signs such that only $(a-b),(b-c),(c-a)$ appear in the denominators of the LHS. Multiply both sides by $(a-b)(b-c)(c-a)$. Expand everything and simplify. In $$ -a^3 (b - c) - b^3 (c - a) - c^3 (a - b) = (a + b + c) (a - b) (b - c) (c - a)$$ both sides are homogeneous polynomial in $a,b,c$ with degree $4$. – Jack D'Aurizio Mar 21 '18 at 22:20