From \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Hence \(A=\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\left(a+b+c=0\right)\)

Done  !!

You're wrong.The question gives a + b + c = 3