Prove $\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0$ if $\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$

Question

if $a,b,c$ are real numbers and $$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$$ Prove $$\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0$$

things i have done: using the assumption i deduced that $$\frac{a}{b-c}=-\frac{(b-c)(a-b-c)}{(c-a)(a-b)}\rightarrow\frac{a}{(b-c)^2}=\frac{(c+b-a)}{(c-a)(a-b)}$$

So some rewriting $$\frac{(c+b-a)}{(c-a)(a-b)}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2} =\frac{(c+b-a)(c-a)(a-b)+b(a-b)^2+a(c-a)^2}{(c-a)^2(a-b)^2}$$

I tried to write numerator in expanded form but it was to large and it seemed like there was not going to be something useful after factoring. so is there any better way than my brute force ? can my solution get continued?

Answer 1

$$\left(\frac a{b-c}+\frac b{c-a}+\frac c{a-b}\right)\left(\frac1{b-c}+\frac1{c-a}+\frac1{a-b}\right)$$

$$=\sum\frac a{(b-c)^2}+\sum\frac b{c-a}\left(\frac1{b-c}+\frac1{a-b}\right)$$

$$=\sum\frac a{(b-c)^2}-\sum\frac b{c-a}\left(\frac{c-a}{(b-c)(a-b)}\right)$$

$$=\sum\frac a{(b-c)^2}-\frac{\sum b(c-a)}{(b-c)(a-b)(c-a)}$$

Now, $\displaystyle\frac{\sum b(c-a)}{(b-c)(a-b)(c-a)}=0$

Answer 2

here is an another (thanks to marty cohen suggestion on comments).

$$\sum\limits_{cyc}\frac{a}{(b-c)^2}=\sum\limits_{cyc}\frac{c+b-a}{(c-a)(a-b)}=\frac{\sum\limits_{cyc}(c+b-a)(b-c)}{(a-b)(b-c)(c-a)}=0$$

Answer 3

Put $$P(x)=(x-b+c)(x-c+a)(x-a+b)=x^{3}+\alpha x^2+\beta x+\gamma$$

We immediately have $\alpha=0$, and $\gamma\not =0$ (as $a,b,c$ are distincts).

Put $w_n=a(b-c)^n+b(c-a)^n+c(a-b)^n$ for $n\in \mathbb{Z}$. We see $w_1=0$ (easy computation) and $w_{-1}=0$ (by hypothesis). Now we have $w_{n+3}+\beta w_{n+1}+\gamma w_n=0$ for all $n\in \mathbb{Z}$. Take $n=-2$ and we get $w_{-2}=0$.