2ab+bc+ca Formula

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$\\Solution:\\\\ Given:\ a+b+c+c=0\\\\\\ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} =\frac{a^3+b^3+c^3}{abc}\\\\\\ We\ know,\\\\ a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc\\\\ \text{If a+b+c=0 then\ } a^3+b^3+c^3=3abc\\\\\\ =\frac{3abc}{abc}\\\\\\ =3$

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$\\ = \frac12\times2(a^2 + b^2 + c^2 - ab - bc - ca)\\\\ = \frac12(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)\\\\ = \frac12(a^2 - 2ab + b^2 + b^2- 2bc + c^2 + c^2 - 2ca + a^2)\\\\ = \frac12 [(a-b)^2+ (b-c)^2+ (c-a)^2]$

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$\\Solution:\\\\ Given:\\\\ a^2+b^2+c^2-ab-bc-ca=0\\\\ \Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)=0\\\\ \Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\\\ \Rightarrow (a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ca)=0\\\\ \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\\\\ Since\ sum\ of\ the\ squares\ are\ zero\ then\\\\ \Rightarrow (a-b)^2=0\Rightarrow a=b\\\\ \Rightarrow (b-c)^2=0\Rightarrow b=c\\\\ \Rightarrow (c-a)^2=0\Rightarrow c=a\\\\ \therefore a=b=c$

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