Thus $$\begin{aligned}
 \label{CMCM}
M&=& \left|
\begin{array}{cccccc}
0&1&1&1& \cdots\cdots & 1\\
1&0&x_{1,2}^2&x_{1,3}^2& \cdots\cdots&x_{1,n+1}^2\\
1&x_{2,1}^2&0&x_{2,3}^2& \cdots\cdots&x_{2,n+1}^2\\
1&x_{3,1}^2&x_{3,2}^2& 0& \cdots\cdots&x_{3,n+1}^2\\
\cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots \\
\cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots \\
\cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots & \cdots \cdots \\
1&x_{n+1,1}^2&x_{n+1,2}^2& x_{n+1,3}^2& 
\cdots \cdots & 0
\end{array}
\right|.\end{aligned}$$ Then

1.  $M$ is homogeneous of homogeneity degree $2n$,

2.  $M$ is reducible if and only if $n=2$. In this case, if we let $$x_{1,2} = z,~x_{2,3} = x,~x_{1,3}=y,$$ then $$\begin{aligned}
    M&=& \left|
    \begin{array}{cccccc}
    0&1&1&1\\
    1&0&z^2&y^2\\
    1&z^2&0&x^2\\
    1&y^2&x^2& 0
    \end{array}
    \right| \nonumber \\ &=& -(x + y + z) (-x+y+z) (x - y + z) (x+y-z). \label{Heron-M}\end{aligned}$$

[*Proof.*]{} (i) Let $M_1$ be the matrix obtained from $M$ by replacing each $x_{i,j}$ in $M$ by $\lambda x_{i,j}$, let $M_2$ be obtained from $M_1$ by pulling out $\lambda^2$ as a common factor in each row of $M_1$ except the upper most one, and let $M_3$ be obtained from $M_2$ by pulling out $1/\lambda^2$ as a common factor in the left most column. Then 