A) \[{{S}_{6}}\]
B) \[{{S}_{5}}-{{S}_{3}}\]
C) \[{{S}_{6}}-{{S}_{4}}\]
D) None
Correct Answer: D
Solution :
[d] \[\Delta \left| \begin{matrix} {{S}_{0}} & {{S}_{1}} & {{S}_{2}} \\ {{S}_{1}} & {{S}_{2}} & {{S}_{3}} \\ {{S}_{2}} & {{S}_{3}} & {{S}_{4}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1+1+1 & \alpha +\beta +\gamma & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \\ \alpha +\beta +\gamma & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}} \\ {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}} & {{\alpha }^{4}}+{{\beta }^{4}}+{{\gamma }^{4}} \\ \end{matrix} \right|\] The above determinant can be expressed as product of two determinants. Thus, \[\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ {{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\ \end{matrix} \right|\left| \begin{matrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ {{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\ \end{matrix} \right|\] \[={{[(\beta -\alpha )(\gamma -\alpha )(\gamma -\beta )]}^{2}}\]
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