If $A=[ij(i+j)]_{n \times n}$, why $\det(A)=0$, when $n>2$ My numerical experiments suggest that if $A=[ij(i+j)]_{n \times n}$, then $\det(A)=0$, $i,j \in[1,n] $ and  $n>2.$
What could be an analytic proof for this observation? Can there be a general result?
EDIT:
Further, if $A=[(i+j)^k]_{n\times n}$, then $\det(A)=0$ for $n\ge 2+k$. The idea of degree of polynomial $k$  seems to work here according to the rank connection: $ rank(A+B+C+...) \le[ rank(A)+ rank(B)+rank(C)+...]$ or the matrices as pointed out by @orangeskid in his answer below.
This is why for $A=[i^k+j^k]_{n\times n}, k=1,2,3,4,5,...$ we have $\det(A)=0$ for $n\ge 3$ (independent of $k$).
 A: HINT:
It's a sum of two matrices of rank $1$, so at rank at most $2$.
A: A general observation is that for real numbers $x_j$ and polynomials $q_1, \dots, q_n$ of degree $k<n-1$, the determinant with entries $q_i(x_j)$ has value zero.
A: It's easily seen that $A$ is the sum of two rank-$1$ matrices. Let's try other ways.
One approach: compute the explicit rank factorization of $A$
Let $n\ge2$, and let $P$ the matrix formed by the first two columns of $A$ and $Q$ the $2\times n$ matrix such that:
$$Q_{1,j}=j(2-j)$$
$$Q_{2,j}=j(j-1)/2$$
If $M=PQ$, then
$$M_{i,j}=i(i+1)\times j(2-j)+2i(i+2)\times j(j-1)/2\\=ij(2i+2-ij-j+ij-i+2j-2)=ij(i+j)$$
That is, $M=A$ and $A=PQ$ is the rank factorization of $A$.
Hence $A$ has rank $2$.

Another way:
Let $B$ the $n\times n$ matrix such that:
$$B_{i,j}=(-1)^{i+j}\binom{j-1}{i-1}$$
And $\det B=1$, since $B$ is an upper-triangular matrix with ones on the diagonal.
Let $C=AB$.
Then up to sign, the $i$-th row of $C$ is the binomial transform of the $i$-th row of $A$:
$$C_{i,j}=\sum_{k=1}^j (-1)^{j+k}\binom{j-1}{k-1}A_{i,k}$$
Now an interesting property of the binomial transform is that the transform of a sequence that is a polynomial of degree $p$ is zero after the first $p+1$ terms.
This is due to the fact that the $j$-th term is the $j$-th discrete difference of the transformed sequence. Taking the index shift into account:
$$|C_{i,j}|=|\Delta^{j-1}A_{i,1}|,$$
where the difference is computed on the column index.
And the $j$-th difference of a polynomial of degree $k$ is a polynomial of degree $k-j$ if $j\le k$, and zero otherwise. It's easy to prove for monomials by induction on $j$, then by linearity it's proved for all polynomials.
That is, if the rows of $A$ are polynomials in $j$ of degree $k$, then $C$ has at most $k+1$ nonzero columns.
And since $\det A=\det C$, if $A$ has dimension greater than $k+1$, then $\det A=0$.
In your case, the rows of $A$ are polynomials of degree $2$ in $j$, so $\det A=0$ for $n>3$. This leaves the case $n=3$ as open, but it's easy to check that the determinant is zero too.

The previous argument is just another way to write that, for any matrix $A$, $\det A=\det C$ with $C_{i,j}=\Delta^{j-1}A_{i,1}$, where the forward difference is computed on the column index. This is due to the linearity of the determinant.
