Given that $a,b,c$ are distinct real numbers such that $a^{2}*(1-b+c)+b^{2}*(1-c-a)+c^{2}*(1-a+b)=ab+bc+ca$  First, I tried to expand the bottom of what we need to prove.  $\frac{1}{a^{2}-2ab+b^{2}}+\frac{1}{b^{2}-2bc+c^{2}}+\frac{1}{c^{2}-2ac+a^{2}}$  I then tried to expand what we are given.  $a^{2}-a^{2}*b+a^{2}*c+b^{2}-b^{2}*c-b^{2}*a+c^{2}-c^{2}*a+c^{2}*b=ab+bc+ca$  $a^{2}-a^{2}*b+a^{2}*c+b^{2}-b^{2}*c-b^{2}*a+c^{2}-c^{2}*a+c^{2}*b-ab-bc-ca=0$  Add 1 to both sides  $a^{2}-a^{2}*b+a^{2}*c+b^{2}-b^{2}*c-b^{2}*a+c^{2}-c^{2}*a+c^{2}*b-ab-bc-ca+1=1$  Help will be appreciated.
$a,b,c$ 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.  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.
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