A) \[abc\]
B) 0
C) \[a+b+c\]
D) \[3\text{ }abc\]
Correct Answer: D
Solution :
Since centroid of triangle formed by \[(a,\,b),\,(b,c)\] and \[(c,\,a)\] is at the origin. \[\therefore \] \[\frac{a+b+c}{3}=0\,\,\,\,\Rightarrow \,\,a+b+c=0\] ?.(i) Also, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\frac{1}{2}[a+b+c]\] \[[{{(a-b)}^{2}}{{(b-c)}^{2}}+{{(c-a)}^{2}}]\] \[\Rightarrow \] \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\] (from (i)) \[\Rightarrow \] \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]You need to login to perform this action.
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