If ${\text{⨍ (n)=α}}^{\mathrm{n}}{\text{+β}}^{\mathrm{n}}$ and $\left|\begin{array}{ccc}3& 1+\text{⨍ (1)}& 1+\text{⨍ (2)}\\ 1+\text{⨍ (1)}& 1+\text{⨍ (2)}& 1+\text{⨍ (3)}\\ 1+\text{⨍ (2)}& \text{1+⨍ (3)}& 1+\text{⨍ (4)}\end{array}\right|=\mathrm{k}{(1-\mathrm{\alpha })}^2 {(1-\mathrm{\beta })}^2 {(\mathrm{\alpha }-\mathrm{\beta })}^2,$ 

then $\mathrm{k}$ is equal to

 

Let $$\begin{gathered} ∆=\left|\begin{array}{ccc}3& 1+\mathrm{\alpha }+\mathrm{\beta }& 1+{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\\ 1+\mathrm{\alpha }+\mathrm{\beta }& 1+{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2& 1+{\mathrm{\alpha }}^3+{\mathrm{\beta }}^3\\ 1+{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2& 1+{\mathrm{\alpha }}^3+{\mathrm{\beta }}^3& 1+{\mathrm{\alpha }}^4+{\mathrm{\beta }}^4\end{array}\right| \\ \Rightarrow ∆=\left|\begin{array}{ccc}1& 1& 1\\ 1& \mathrm{\alpha }& \mathrm{\beta }\\ 1& {\mathrm{\alpha }}^2& {\mathrm{\beta }}^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \mathrm{\alpha }& \mathrm{\beta }\\ 1& {\mathrm{\alpha }}^2& {\mathrm{\beta }}^2\end{array}\right|{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1;{\mathrm{C}}_3\to {\mathrm{C}}_2-{\mathrm{C}}_1 \\ \Rightarrow ∆={\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -1\\ 1& {\alpha }^2-1& {\beta }^2-1\end{array}\right|}^2={\left(\alpha -1\right)}^2{\left(\beta -1\right)}^2\left|\begin{array}{cc}1& 1\\ \alpha +1& \beta +1\end{array}\right| \\ \Rightarrow ∆={\left(\alpha -1\right)}^2{\left(\beta -1\right)}^2{\left(\alpha -\beta \right)}^2={\left(1-\alpha \right)}^2{\left(1-\beta \right)}^2{\left(\alpha -\beta \right)}^2 \end{gathered}$$