A) \[a=-b=-c\]
B) \[a=2b=2c\]
C) \[a=b=c\],\[a=-2b=-2c\]
D) None of these
Correct Answer: C
Solution :
As \[f(x)={{x}^{3}}-3{{b}^{2}}x+2{{c}^{3}}\] is divisible by \[x-a\] and \[x-b\], therefore \[f(a)=0\,\,\Rightarrow {{a}^{3}}-3{{b}^{2}}a+2{{c}^{3}}=0\] .....(i) and \[f(b)=0\]Þ\[{{b}^{3}}-3{{b}^{3}}+2{{c}^{3}}=0\] .....(ii) From (ii), \[b=c\] From (i), \[{{a}^{3}}-3a{{b}^{2}}+2{{b}^{3}}=0\](Putting \[b=c\]) \[\Rightarrow (a-b)({{a}^{2}}+ab-2{{b}^{2}})=0\] \[\Rightarrow \]\[a=b\] or \[{{a}^{2}}+ab=2b\] Thus \[a=b=c\] or \[{{a}^{2}}+ab=2{{b}^{2}}\] and \[b=c\] \[{{a}^{2}}+ab=2{{b}^{2}}\]is satisfied by \[a=-2b\]. But \[b=c\]. \\[{{a}^{2}}+ab-2{{b}^{2}}\] and \[b=c\] is equivalent to \[a=-2b=-2c\]
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