A) \[{{y}^{3}}+y+2=0\]
B) \[{{y}^{3}}-{{y}^{2}}-y-2=0\]
C) \[{{y}^{3}}+3{{y}^{2}}-y-3=0\]
D) \[{{y}^{3}}+4{{y}^{2}}+5y+20=0\]
Correct Answer: B
Solution :
Given equation\[{{x}^{3}}-3{{x}^{2}}+x+5=0\]. Then\[\alpha +\beta +\gamma =3\], \[\alpha \beta +\beta \gamma +\gamma \alpha =1\], \[\alpha \beta \gamma =-5\] \[y=\Sigma {{\alpha }^{2}}+\alpha \beta \gamma ={{(\alpha +\beta +\gamma )}^{2}}-2\,(\alpha \beta +\beta \gamma +\gamma \alpha )+\alpha \beta \gamma \] = \[9-2-5=2\] \[\therefore \] \[y=2\] It satisfies the equation\[{{y}^{3}}-{{y}^{2}}-y-2=0\].
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