Ab+bc+ca0

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$\\ The \ given \ equation\ is \ $(b-c) x^{2}+(c-a) x+(a-b)=0$ \\ The \ roots \ of \ the \ given \ equation \ are \ equal, \\ \\ $D=b^{2}-4 a c=0$ \\ $\Rightarrow(c-a)^{2}-4(b-c)(a-b)=0$ \\ $\Rightarrow\left(c^{2}+a^{2}-2 a c\right)-4\left(a b-b^{2}-c a+b c\right)=0$ \\ $\Rightarrow c^{2}+a^{2}-2 a c-4 a b+4 b^{2}+4 a c-4 b c=0$ \\ $\Rightarrow a^{2}+4 b^{2}+c^{2}-4 a b-4 b c+2 a c=0$ \\ $\Rightarrow a^{2}+(-2 b)^{2}+c^{2}+2 \times a \times(-2 b)+2 \times(-2 b) \times c+2 \times c \times a=0$ \\ $\Rightarrow(a+(-2 b)+c)^{2}=0 \\ \quad\left(\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x\right)$ \\ $\Rightarrow a-2 b+c=0$ \\ $\Rightarrow 2 b=a+c$$

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

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Ab+bc+ca0

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