If α, β ≠ 0 , f(n) = α ^ n + β ^ n and | cases 3 & 1 + f(1) & 1 +...

f\left(1\right) = \alpha + \beta, f\left(2\right) = {\alpha}^{2} + {\beta}^{2}, f\left(3\right) = {\alpha}^{3} + {\beta}^{3}, f\left(4\right) = {\alpha}^{4} + {\beta}^{4}</p><p>So, the given determinant can be written as<br /><br />\left|\begin{cases}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{cases}\right| = \left|\begin{cases}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha}^{2} & {\beta}^{2}\end{cases}\right|\left|\begin{cases}1 & 1 & 1 \\ 1 & \alpha & {\alpha}^{2} \\ 1 & \beta & {\beta}^{2}\end{cases}\right|

= \left|\begin{cases}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha}^{2} & {\beta}^{2}\end{cases}\right| \left|\begin{cases}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha}^{2} & {\beta}^{2}\end{cases}\right|

= {\left|\begin{cases}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha}^{2} & {\beta}^{2}\end{cases}\right|}^{2} = {\left[\left(\alpha {\beta}^{2} - {\alpha}^{2} \beta\right) - \left({\beta}^{2} - \beta\right) + \left({\alpha}^{2} - \alpha\right)\right]}^{2}

$= {\left[\alpha \beta \left(\beta - \alpha\right) - \left(\beta - \alpha\right) \left(\beta + \alpha\right) + \left(\beta - \alpha\right)\right]}^{2}$

$= {\left(\beta - a\right)}^{2}{\left[\alpha \beta - \beta - \alpha + 1\right]}^{2}$

$= {\left(\beta - \alpha\right)}^{2} {\left(\alpha - 1\right)}^{2} {\left(\beta - 1\right)}^{2}$

Hence, $K = 1$.