Question

Let $S_{k} = \alpha^{k} + \beta^{k} + \gamma^{k},$ then
$\Delta =\begin{vmatrix}S_{0}&S_{1}&S_{2}\\ S_{1}&S_{2}&S_{3}\\ S_{2}&S_{3}&S_{4}\end{vmatrix}$ is equal to

• A. $S_6$
• B. $S_5 - S_3$
• C. $S_6 - S_4$
• D. None of these

Answer 1

Answer: (D)

$\Delta =\begin{vmatrix}S_{0}&S_{1}&S_{2}\\ S_{1}&S_{2}&S_{3}\\ S_{2}&S_{3}&S_{4}\end{vmatrix}$
$= \begin{vmatrix}1+1+1&\alpha+\beta+\gamma&\alpha ^{2}+\beta^{2} +\gamma ^{2}\\ \alpha +\beta +\gamma &\alpha ^{2}+\beta ^{2} +\gamma ^{2}&\alpha ^{3}+\beta ^{3} +\gamma ^{3}\\ \alpha ^{2}+\beta ^{2} +\gamma ^{2}&\alpha ^{3}+\beta ^{3} +\gamma ^{3}&\alpha ^{4}+\beta ^{4} +\gamma ^{4}\end{vmatrix}$
The above determinant can be expressed as product of two determinants. Thus,
$\Delta = \begin{vmatrix}1&1&1\\ \alpha&\beta&\gamma\\ \alpha^{2}&\beta^{2}&\gamma^{2}\end{vmatrix}\begin{vmatrix}1&1&1\\ \alpha &\beta &\gamma \\ \alpha ^{2}&\beta ^{2}&\gamma ^{2}\end{vmatrix} = \left[\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)\right]^{2}$