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if a+b+c = 0 then find (a+b)²/ ab + (b+c)²/bc + (c+a)²/ca

\frac{{(a + b)}^{2}}{ab} + \frac{{(b + c)}^{2}}{bc} + \frac{{(c + a)}^{2}}{ca} = \frac{{(- c)}^{2}}{ab} + \frac{{(- a)}^{2}}{bc} + \frac{{(- b)}^{2}}{ca} = \frac{c^{2}}{ab} + \frac{a^{2}}{bc} + \frac{b^{2}}{ca} = \frac{c^{3} + a^{3} + b^{3}}{abc} If a + b + c = 0 then a^{3} + b^{3} + c^{3} = 3 abc So, \frac{c^{3} + a^{3} + b^{3}}{abc} = \frac{3 abc}{abc} = 3 ANS . . .

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