Given,

$A={\left[a_{ij}\right]}_{m\times n}$

$a_{ij}=\left(i^2+j^2-ij\right)\left(j-i\right)$

Now, Replace $i\to j, j\to i$

$$\begin{gathered} a_{ji}=\left(i^2+j^2-ij\right)\left(i-j\right) \\ \Rightarrow a_{ji}=-\left(i^2+j^2-ij\right)\left(j-i\right) \\ \Rightarrow a_{ji}=-a_{ij} \end{gathered}$$

It is a skew-symmetric matrix where $a_{ij}+a_{ji}$ is equal to zero.

The determinant of Skew-Symmetric matrix of odd order$\left(n=3\right)$ is always zero.

$\left|A\right|=0$

All diagonal elements of Skew-Symmetric matrix are zero.

So, 
$\mathrm{Tr}\left(A\right)=\sum _{i=1}^n{a}_{ii}=0$