A) \[{{x}^{3}}+1\]
B) \[{{x}^{3}}+\omega \]
C) \[{{x}^{3}}+{{\omega }^{2}}\]
D) \[{{x}^{3}}\]
Correct Answer: D
Solution :
\[\Delta =\left| \,\begin{matrix} x+1 & \omega & {{\omega }^{2}} \\ \omega & x+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & x+\omega \\ \end{matrix}\, \right|\] =\[\left| \,\begin{matrix} x+1+\omega +{{\omega }^{2}} & \omega & {{\omega }^{2}} \\ x+1+\omega +{{\omega }^{2}} & x+{{\omega }^{2}} & 1 \\ x+1+\omega +{{\omega }^{2}} & 1 & x+\omega \\ \end{matrix}\, \right|\],\[({{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}})\] = \[x\,\left| \,\begin{matrix} 1 & \omega & {{\omega }^{2}} \\ 1 & x+{{\omega }^{2}} & 1 \\ 1 & 1 & x+\omega \\ \end{matrix}\, \right|\] \[\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|\ne 0\] = \[x\,[1\{(x+{{\omega }^{2}})\,(x+\omega )-1\}+\omega \{1-(x+\omega )\}\] \[+{{\omega }^{2}}\{1-(x+{{\omega }^{2}})\}]\] = \[x({{x}^{2}}+\omega x+{{\omega }^{2}}x+{{\omega }^{3}}-1+\omega -\omega x-{{\omega }^{2}}\]\[+{{\omega }^{2}}-{{\omega }^{2}}x-{{\omega }^{4}})\] = \[{{x}^{3}}\] , \[(\because \,\,{{\omega }^{3}}=1)\].
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