A) \[3(a-b)(b-c)(c-a)\]
B) \[3(a+b)(b+c)(c+a)\]
C) \[(b-a)(b-c)(c-a)\]
D) \[(a+b)(b+c)(c+a)\]
Correct Answer: B
Solution :
If\[a+b+c=0\] \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\] Here, \[{{a}^{2}}-{{b}^{2}}+{{b}^{2}}-{{c}^{2}}+{{c}^{2}}-{{a}^{2}}=0\] and \[a-b+b-c+c-a=0\] \[\therefore \]Expression \[=\frac{3({{a}^{2}}-{{b}^{2}})({{b}^{2}}-{{c}^{2}})({{c}^{2}}-{{a}^{2}})}{(a-b)(b-c)(c-a)}\] \[=2(a+b)(b+c)(c+a)\]You need to login to perform this action.
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