A) \[k(a+b)(b+c)(c+a)\]
B) \[k\,abc({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]
C) \[k(a-b)(b-c)(c-a)\]
D) \[k(a+b-c)(b+c-a)(c+a-b)\]
Correct Answer: C
Solution :
[c] we have, \[=k(a-b)(b-c)(c-a)\]You need to login to perform this action.
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