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

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