If a+b+c=0, then the value of left(frac{a^{2} | vedclass

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(D) Given that $a+b+c=0.$We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).$Since $a+b+c=0,$ the right sid

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$3$

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(D) Given that $a+b+c=0.$We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).$Since $a+b+c=0,$ the right side becomes $0,$ so $a^{3}+b^{3}+c^{3}=3abc.$Now,consider the expression: $\frac{a^{2}}{bc} + \frac{b^{2}}{ca} + \frac{c^{2}}{ab}.$Taking the least common multiple $(LCM)$ as $abc,$ we get: $\frac{a^{3}+b^{3}+c^{3}}{abc}.$Substituting $a^{3}+b^{3}+c^{3}=3abc$ into the expression,we get: $\frac{3abc}{abc} = 3.$

If $a+b+c=0,$ then the value of $\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right)$ is:

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