A) \[a=-b=-c\]
B) \[a=2b=2c\]
C) \[4{{b}^{2}}=ac\]
D) None of these
Correct Answer: D
Solution :
As\[f(x)~={{x}^{3}}-3{{b}^{2}}x+2{{c}^{2}}\] is divisible by\[(x-a)\] and\[(x-b)\]. Now, \[f(a)=0\] \[\Rightarrow \] \[{{a}^{3}}-3{{b}^{2}}a+2{{c}^{3}}=0\] ...(i) and \[f(b)=0\] \[\Rightarrow \] \[{{b}^{3}}-3{{b}^{3}}+2{{c}^{3}}=0\] ...(ii) From Eq. (ii), \[b=c\] On putting\[b=c\]in Eq. (i), we get \[{{a}^{3}}-3a{{b}^{2}}+2{{b}^{3}}=0\] \[\Rightarrow \] \[(a-b)({{a}^{2}}+ab-2{{b}^{2}}\}=0\] \[\Rightarrow \] \[a=b\] or \[{{a}^{2}}+ab=2{{b}^{2}}\] Thus, \[a=b=c\] or\[{{a}^{2}}+ab=2{{b}^{2}}\] and \[b=c\] Therefore, \[{{a}^{2}}+ab=2{{b}^{2}}\] and \[b=c\] is equivalent to \[a=-2b=-2c\]
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