question_answer

If \[f(x)=a+bx+c{{x}^{2}}\] and \[\alpha ,\beta ,\gamma \]are the roots of the equation\[{{x}^{3}}=1,\]then is equal to

A)  \[f(\alpha )+f(\beta )+f(\gamma )\]

B)  \[f(\alpha )f(\beta )+f(\beta )\]\[f(\gamma )+f(\gamma )\]\[f(\alpha )\]

C)  \[f(\alpha )f(\beta )f(\gamma )\]

D)  \[f(\alpha )f(\beta )f(\gamma )\]

Correct Answer: B

Solution :

[d] =-({{a}^{3}}+{{b}^{3}}+{{c}^{3}}-abc)\] \[=-(a+b+c)(a+b{{\omega }^{2}}+c\omega )\] \[(a+b\omega +c{{\omega }^{2}})\] (Where \[\omega \]is cube roots of unity) \[[\therefore \alpha =1,\beta =\omega ,\gamma ={{\omega }^{2}}]\] \[=-f(\alpha )f(\beta )f(\gamma )\]