A) 3
B) 2
C) 0
D) 1
Correct Answer: C
Solution :
[c] \[a=1,\]\[b=2\]and \[c=-\,3\] Now, when a + b + c = 0 \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] Here, \[a+b+c=(1+2-3)=0\] \[\therefore \] \[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{ab+bc+ca-({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\] \[=\frac{3abc-3abc}{ab+bc+ca-({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}=0\] |
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