Question

Let $S_k={\alpha }^k+{\beta }^k+{\gamma }^k,$ then
$\Delta =\left|\begin{array}{ccc}{S}_0& S_1& S_2\\ S_1& S_2& S_3\\ S_2& S_3& S_4\end{array}\right|$ is equal to

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

Answer 1

Answer: (D)

$\Delta =\left|\begin{array}{ccc}{S}_0& S_1& S_2\\ S_1& S_2& S_3\\ S_2& S_3& S_4\end{array}\right|$
$=\left|\begin{array}{ccc}1+1+1& \alpha +\beta +\gamma & {\alpha }^2+{\beta }^2+{\gamma }^2\\ \alpha +\beta +\gamma & {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^3+{\beta }^3+{\gamma }^3\\ {\alpha }^2+{\beta }^2+{\gamma }^2& {\alpha }^3+{\beta }^3+{\gamma }^3& {\alpha }^4+{\beta }^4+{\gamma }^4\end{array}\right|$
The above determinant can be expressed as product of two determinants. Thus,
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ \alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ \alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\end{array}\right|={\left[\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left(\gamma -\beta \right)\right]}^2$