Simil-Vandermonde determinant Compute the determinant:
$$ \det\begin{bmatrix}
1      & x_{1}  & x_{1}^{2}& \dots & x_{1}^{n-2} & (x_{2}+ x_{3}+ \dots + x_{n})^{n-1}  \\
1      & x_{2}  & x_{2}^{2}& \dots & x_{2}^{n-2} & (x_{1}+ x_{3}+ \dots + x_{n})^{n-1}  \\
\vdots & \vdots &\vdots &\dots & \vdots& \vdots \\
1      & x_{n}  & x_{n}^{2}& \dots & x_{n}^{n-2} & (x_{1}+ x_{2}+ \dots + x_{n-1})^{n-1}  \\ 
\end{bmatrix}.$$

Put $$ A(x_1,\ldots,x_n)=\begin{bmatrix}
1      & x_{1}  & x_{1}^{2}& \dots & x_{1}^{n-2} & (x_{2}+ x_{3}+ \dots + x_{n})^{n-1}  \\
1      & x_{2}  & x_{2}^{2}& \dots & x_{2}^{n-2} & (x_{1}+ x_{3}+ \dots + x_{n})^{n-1}  \\
\vdots & \vdots &\vdots &\dots & \vdots& \vdots \\
1      & x_{n}  & x_{n}^{2}& \dots & x_{n}^{n-2} & (x_{1}+ x_{2}+ \dots + x_{n-1})^{n-1}  \\ 
\end{bmatrix}.$$
Then we have 
\begin{align*}
\det A(x_1,\ldots,x_n)&= \begin{vmatrix}
1      & x_{1}  & x_{1}^{2}& \dots & x_{1}^{n-2} & (x_{2}+ x_{3}+ \dots +\ x_{n})^{n-1}  \\
1      & x_{2}  & x_{2}^{2}& \dots & x_{2}^{n-2} & (x_{1}+ x_{3}+ \dots +\ x_{n})^{n-1}  \\
\vdots & \vdots &\vdots&&\vdots& \vdots \\
1      & x_{n}  & x_{n}^{2}& \dots & x_{n}^{n-2} & (x_{1}+ x_{2}+ \dots +\ x_{n-1})^{n-1} 
\end{vmatrix}\\
&=\begin{vmatrix}
1      & x_{1}  & x_{1}^{2}& \dots & x_{1}^{n-2} & \sum_{k=0}^{n-1}\binom {n-1}k(-1)^kx_1^k(x_1+\ldots + x_n)^{n-k-1} \\
1      & x_{2}  & x_{2}^{2}& \dots & x_{2}^{n-2} &\sum_{k=0}^{n-1}\binom {n-1}k(-1)^kx_2^k(x_1+\ldots +x_n)^{n-k-1} \\
\vdots & \vdots &\vdots&&\vdots& \vdots \\
1      & x_{n}  & x_{n}^{2}& \dots & x_{n}^{n-2} & \sum_{k=0}^{n-1}\binom {n-1}k(-1)^kx_n^k(x_1+\ldots  + x_n)^{n-k-1}
\end{vmatrix}\\
&=\sum_{k=0}^{n-1}\binom{n-1}k(-1)^k(x_1+\ldots +x_n)^{n-k-1}\begin{vmatrix}
1      & x_{1}  & x_{1}^{2}& \dots & x_{1}^{n-2} & x_1^k \\
1      & x_{2}  & x_{2}^{2}& \dots & x_{2}^{n-2} &x_2^k \\
\vdots & \vdots&\vdots &&\vdots& \vdots \\
1      & x_{n}  & x_{n}^{2}& \dots & x_{n}^{n-2} &x_n^k
\end{vmatrix}\\
&=(-1)^{n-1}\begin{vmatrix}
1      & x_{1}  & x_{1}^{2}& \dots & x_{1}^{n-2} & x_1^{n-1} \\
1      & x_{2}  & x_{2}^{2}& \dots & x_{2}^{n-2} &x_2^{n-1} \\
\vdots & \vdots &\vdots&&\vdots& \vdots \\
1      & x_{n}  & x_{n}^{2}& \dots & x_{n}^{n-2} &x_n^{n-1}
\end{vmatrix}
\end{align*}
hence
$$\det A(x_1,\ldots,x_n)=(-1)^{n-1}\prod_{1\leq i<j\leq n}(x_j-x_i).$$
