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6 years ago


Answers : (1)


Dear ajinkya

a2(b-c)3+b2(c-a)3+c2(a-b)3   = ∑a2(b-c)3

                                                      ={a2b3-a2c3 -3abc(ab-ac)}

                                                      ={a2b3-a2c3) -3abc(ab-ac)

                                                      ={a2b3-a2c3) -0

                                                       =a2b3-a2c+b2c3-b2a3   + c2a3-c2b3

                                                        =a2b2(b-a) +b2c2(c-b) + c2a2(a-c)

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