An inequality with a weird condition $a,b,c$ are distinct real numbers that $$(a^2)(1-b+c)+(b^2)(1-c+a)+(c^2)(1-a+b)=ab+bc+ca.$$ Prove $$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.$$
I have tried two different approaches to this problem but am stuck.

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*First I started by expanding the bottom of each fraction.
$\frac{1}{(a^2)-2ab+(b^2)}+\frac{1}{(b^2)-2bc+(c^2)}+\frac{1}{(c^2)-2ac+(a^2)}=1$


*I also tried to expand $(a^2)(1-b+c)+(b^2)(1-c-a)+(c^2)(1-a+b)=ab+bc+ca$
$(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b=ab+bc+ca$
$(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca=0$
Then I added one to both sides
$(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca+1=0+1$
Help will be appreciated.
 A: Let $t=a-b,\; u=b-c,\; v=c-a$,$\;\;$ so $t+u+v=0$ and $v=-(t+u)$.
Then $\displaystyle\sum_{cyc}ab=\sum_{cyc}a^2(1-b+c)\implies\sum_{cyc}(a^2b-a^2c)=\sum_{cyc}(a^2-ab)\implies$
$(a-b)(b-c)(a-c)=(a-b)^2+(a-b)(b-c)+(b-c)^2\implies tu(t+u)=t^2+tu+u^2$
$\implies t^2u^2(t+u)^2=(t^2+tu+u^2)^2=(t^2+u^2)(t+u)^2+t^2u^2=u^2(t+u)^2+t^2(t+u)^2+t^2u^2$.
Dividing both sides by $t^2u^2(t+u)^2$ gives $\displaystyle1=\frac{1}{t^2}+\frac{1}{u^2}+\frac{1}{(t+u)^2}$, 
so $t+u=-v\implies \displaystyle\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.$
A: Hint: I'll assume the second factor of term two is $(1-c+a)$ as one would expect in such cyclic expressions. If now one considers the left side $w$ of the thing to be shown equal to $1$ (i.e. the sum of the reciprocal squares of differences), it factors nicely as
$$w=\frac{(a^2+b^2+c^2-ab-ac-bc)^2}{[(a-b)(b-c)(c-a)]^2}.$$
So $w=1$ iff the unsquared top (call that $T$) of it is $\pm$ its unsquared bottom (call that $B$).
But if one expands the assumed equation, after moving the right side over, and compares with the sum $T+B$ or difference $T-B$ there's a match which I think shows what you want. I haven't completely spelled it out, hence called it a hint.
