A) 1
B) 0
C) - 1
D) 3
Correct Answer: D
Solution :
Since \[a+b+c=0\] or \[a+b=-c\] or \[{{(a+b)}^{3}}=-{{c}^{3}}\] or \[{{a}^{3}}+{{b}^{3}}+3ab(a+b)=-{{c}^{3}}\] or \[{{a}^{3}}+{{b}^{3}}+3b(-c)=-{{c}^{3}}\] or \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\] Dividing both the sides by\[abc\], we get \[\frac{{{a}^{2}}}{bc}+\frac{{{b}^{2}}}{ac}+\frac{{{c}^{2}}}{ab}=3\]You need to login to perform this action.
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