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If, A+B+C=0, then \({{A}^{3}}+{{B}^{3}}+{{C}^{3}}=3ABC\) For numerator, \(A={{a}^{2}}-{{b}^{2}} B={{b}^{2}}-{{c}^{2}} C={{c}^{2}}-{{a}^{2}}\) Add them A+B+C=0  For denominator, A=a-b   B=b-c    C=c-a  Add them,   A+B+C=0

So,

\(\frac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{a^3-b^3-3a^2b+3ab^2+b^3-c^3-3b^2c+3bc^2-c^3-a^3-3c^2a+3ca^2}\\ =\frac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a-b)(b-c)(c-a)}\\ =\frac{(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)}{(a-b)(b-c)(c-a)}\\ = (a+b)(b+c)(c+a)\)

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