Sign of determinant of a $3 \times 3$ matrix with entries $1+\alpha^{i+j-2}+\beta^{i+j-2}$, for distinct $\alpha,\beta\in\mathbb R\setminus\{1\}$

Question

Let $ \alpha\ne1,\beta\ne1$ be the distinct real roots of the equation $$ax^2+bx+c=0,~~a,b,c\in \mathbb{R},a\ne 0$$ Let $S_n=\alpha^n+\beta^n,n\geq0$ and $$\Delta=\begin{vmatrix}1+S_0&1+S_1&1+S_2\\1+S_1&1+S_2&1+S_3\\1+S_2&1+S_3&1+S_4\end {vmatrix}$$then

(A)$\Delta\leq0$

(B)$\Delta>0$

(C)$\Delta<0$

(D)$\Delta=0$

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Here $S_0=\alpha^0+\beta^0,S_1=\alpha^1+\beta^1,S_2=\alpha^2+\beta^2,S_3=\alpha^3+\beta^3,S_4=\alpha^4+\beta^4$

$S_0=2,S_1=\frac{-b}{a},S_2=\frac{b^2-2ac}{a^2},S_3=\frac{-b^3}{a^3}+\frac{3abc}{a^3},S_4=\frac{b^4-6a^2c^2+4bca^2}{a^4}$

If I put these values and simplify, it becomes a very complicated expression,difficult to simplify and judge the sign of $\Delta$. Is there some other method possible, so that I can easily determine the sign of $\Delta$?

Answer 1

Hint: Let $$A:= \begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{bmatrix} $$ then calculate $AA^T$ to see what you get.

Answer 2

Hint see Vandermonde's determinant or write it as a multiplication of two determinants