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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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$\hspace{-16}$The Equation $\bf{x^3-3x+1=0}$ has a Roots $\bf{a\;,b\;,c}$\\\\\\ Then Using Vieta,s Formula\\\\\\ $\bf{a+b+c=0}$\\\\\\ $\bf{ab+bc+ca=-3}$\\\\\\ $\bf{abc=-1}$\\\\\\ If $\bf{a+b+c=0\Leftrightarrow a^3+b^3+c^3=3abc}$\\\\\\ So $\bf{a^3+b^3+c^3=-3\Leftrightarrow \sum_{cyclic(a,b,c)}a^3=-3}$\\\\\\ Similarly $\bf{(a+b+c)^2=\sum_{cyclic(a,b,c)}a^2+2(ab+bc+ca)}$\\\\\\ So $\bf{\sum_{cyclic(a,b,c)}a^2=6}$\\\\\\ Now Given $\bf{x^3=3x-1}$\\\\\\ So $\bf{x^5=3x^3-x^2}$\\\\\\$

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$\\Solution\\\\ Considering\ LHS\\\\ =(a+b)^3+c^3-a^3-b^3-c^3\\\\ =(a+b)^3+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3\ \ \ \ \ \ [(a+b)^3=a^3+b^3+3ab(a+b)]\\\\ =a^3+b^3+3ab(a+b)+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3\\\\ =a^3+b^3+c^3+3ab(a+b)+3c(a+b)(a+b+c)-a^3-b^3-c^3\\\\ =3ab(a+b)+3c(a+b)(a+b+c)\\\\ = 3(a+b)[ab+c(a+b+c)]\\\\ =3(a+b)[ab+ac+bc+c^2] \\\\ =3(a+b)[a(b+c)+c(b+c)]\\\\ =3(a+b)(b+c)(c+a)=RHS\\\\ Hence\ proced$

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