A) \[{{y}^{3}}\]
B) \[{{y}^{3}}-1\]
C) \[y({{y}^{2}}-1)\]
D) \[y({{y}^{2}}-3)\]
Correct Answer: A
Solution :
Roots of the equation \[{{x}^{2}}+x+1=0\]are \[\alpha =\omega \] and \[\beta ={{\omega }^{2}}\] where \[\omega ,{{\omega }^{2}}\]are complex cube roots of unity \[\therefore \]\[\Delta =\left| \begin{matrix} y+1 & \omega & {{\omega }^{2}} \\ \omega & y+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & y+\omega \\ \end{matrix} \right|\] \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] \[\Rightarrow \Delta =y\left| \begin{matrix} 1 & 1 & 1 \\ \omega & y+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & y+\omega \\ \end{matrix} \right|\] Expanding along \[{{R}_{1}},\]we get\[\Delta =y.{{y}^{2}}\Rightarrow D={{y}^{3}}\]
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