A) \[{{y}^{3}}+y+2=0\]
B) \[{{y}^{3}}-{{y}^{2}}-y-2=0\]
C) \[{{y}^{3}}+3{{y}^{2}}-y-3=0\]
D) \[{{y}^{2}}+4{{y}^{2}}+5y+20=0\]
Correct Answer: B
Solution :
Given \[\alpha ,\beta ,\gamma \] are the roots of equation \[{{x}^{3}}-3{{x}^{2}}+x+5=0\] \[\therefore \] \[\alpha +\beta +\gamma =3\] \[\alpha \beta +\beta \gamma +\gamma \alpha =1\] and \[\alpha \beta \gamma =-5\] Now, \[y=\Sigma {{\alpha }^{2}}+\alpha \beta \gamma \] \[={{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+\alpha \beta \gamma \] \[={{(\alpha +\beta +\gamma )}^{2}}-2(\alpha \beta +\beta \gamma +\gamma \alpha )+\alpha \beta \gamma \] \[={{(3)}^{2}}-2(1)-5\] \[\Rightarrow \] \[y=2\] So, \[y=2\] satisfies the equation \[{{y}^{3}}-{{y}^{2}}-y-2=0\]
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