Question

if a+b+c=0, then prove that a²b²+b²c²+c²a²=(ab+bc+ca)²​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given

a+b+c = 0 ..(1) To prove : a*+b*+c* = 2(a²b²+b²°c²+a²c?) (a+b+c)? = a?+b²+c²+2ab+2bc+2ca (0?) = a?+b?+c²+2ab+2bc+2ca From (1) Or, a?+b?+c? = - 2(ab+bc+ca)...(2) On squaring both sides of (2), we get = a*+b*+c*+2a'b²+2b?c?+2c?a? = 4a b?+4b?c?+4c²a?+8ab°c+8abc²+8a'bc = a*+b*+c* = 4a’b²+4b°c²+4c°a²+8ab?c+ 8abc?+8a*bc-(2a°b²+2b?c²+2c°a?) = a*+b*+c* = 4a’b²+4b?c²+4c?a²+8ab?c+ 8abc?+8a bc-2a'b²-2b°c²-2c?a? > a*+b*+c* = 2a b?+2b?c?+2c²a?+8ab?c+8abc²+8a*bc = a*+b*+c* = 2(a?b²+b²c²+c²a³) +8abc(b+c+a) = a*+b*+c* = 2(a?b²+b°c?+c?a²) +8(0) ..From (1) = a*+b*+c* = 2(a²b²+b²c²+c²a³) Hence proved.