How to solve the given determinant by using properties if possible Let $\alpha,\beta$ be the real roots of the equation $ax^2+bx+c=0$ and let $ {\alpha}^n+{\beta}^n=S_{n}\ for \ n \geq 1 $ then find the value of the determinant of A, where $$A = \left[ \begin{array} { l l l } {\ \ \ \ { 3 }} & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 1 + S _ { 1 } } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ { 1 + S _ { 2 } } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$$
My try:
$A = \left[ \begin{array} { l l l } { 3+0 } & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 1 + S _ { 1 } } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ { 1 + S _ { 2 } } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$
$A = \left[ \begin{array} { l l l } { 3 } & { 1 + S_{1} } & { 1 + S _ { 2 } } \\ {  S _ { 1 } } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ {S _ { 2 } } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]+\left[\begin{array} { l l l } { 0 } & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 1 } & { 1 + S _ { 2 } } & { 1 + S _ { 3 } } \\ { 1  } & { 1 + S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$
Now solving the determinant second by appling operation $R _ { 2 } \rightarrow R _ { 2 } - R _ { 3 }$ we get
$\left[ \begin{array} { l l l } { 0 } & { 1 + S _ { 1 } } & { 1 + S _ { 2 } } \\ { 0 } & { S _ { 2 } - S _ { 3 } } & { S _ { 3 } - S _ { 4 } } \\ { 1 } & { 1 - S _ { 3 } } & { 1 + S _ { 4 } } \end{array} \right]$
$\Rightarrow ( 1 + S _ { 1 } ) ( S _ { 3 } - S _ { 4 } ) - ( S _ { 2 } - S _ { 3 } ) ( 1 + S _ { 2 } )$
But how to solve  other determinant? I am stuck with this question. 
Can anyone give me hint to solve this typical determinant?

$S _ { n } = \alpha ^ { n } + \beta ^ { n } \forall n \geq 1$
$A= \left| \begin{array} { l l l } { 1+1+1 } & { 1+\alpha+\beta } & { 1+\alpha^{2}+\beta^{2} } \\ { 1+\alpha +\beta } & { 1+\alpha^{2}+\beta^{2} } & {  1+\alpha^{3}+\beta^{3}} \\ { 1+\alpha^{2}+\beta^{2} } & { 1+\alpha^{3}+\beta^{3} } & { 1+\alpha^{4}+\beta^{4}} \end{array} \right|$
Now by property of multiplication of determinant we get
$= \left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { \alpha } & { \beta } \\ { 1 } & { \alpha ^ { 2 } } & { \beta ^ { 2 } } \end{array} \right|  \left| \begin{array} { l l l } { 1 } & { 1 } & { 1 } \\ { 1 } & { \alpha } & { \beta } \\ { 1 } & { \alpha ^ { 2 } } & { \beta ^ { 2 } } \end{array} \right|$
$= ( \alpha - \beta ) ^ { 2 } ( \beta - 1 ) ^ { 2 } ( \alpha - 1 ) ^ { 2 }$
