Let $a,b,c,d,e \in \mathbb{R}^3$, and $(w)_{\mathsf{x}}$ denotes the skew matrix associated with the cross product with a three dimensional vector, i.e., for $w,y \in \mathbb{R}^3$, we have $(w)_{\mathsf{x}}y = w \times y$.
Now, consider the following equation:
$$ \square = -(a - b - c)_{\mathsf{x}}(I + (d)_{\mathsf{x}}) + (I + (d)_{\mathsf{x}})(a - e)_{\mathsf{x}} $$
where $I \in \mathbb{R}^{3 \times 3}$ is an identity matrix. Also, from the problem under investigation, $g = e - b$. I've tried a few times to expand this equation to get the answer provided by the author of the paper. Any help will be appreciated. The expected answer, assuming it is correct, is:
$$ \square = (-(a - g)_{\mathsf{x}} d - g + c)_{\mathsf{x}} $$
Edit 1: Below it is what I did so far. I know that $(a)_{\mathsf{x}}(b)_{\mathsf{x}} - (b)_{\mathsf{x}}(a)_{\mathsf{x}} = (a \times b)_{\mathsf{x}} = (a)_{\mathsf{x}}b$. However, I didn't find properties to expand the red part.:
\begin{align} \square &= -(a - b - c)_{\mathsf{x}} -(a - b - c)_{\mathsf{x}} (d)_{\mathsf{x}} + (a-e)_{\mathsf{x}} + (d)_{\mathsf{x}}(a-e)_{\mathsf{x}} \\ \square &= (-a + b + c +a -e)_{\mathsf{x}} -(a - b - c)_{\mathsf{x}} (d)_{\mathsf{x}} + (d)_{\mathsf{x}}(a-e)_{\mathsf{x}}\\ \square &= (b+c-e)_{\mathsf{x}} -(a - b - c)_{\mathsf{x}} (d)_{\mathsf{x}} + (d)_{\mathsf{x}}(a-e)_{\mathsf{x}}\\ \square &= (-g+c)_{\mathsf{x}} \color{red}{-(a - b - c)_{\mathsf{x}} (d)_{\mathsf{x}} + (d)_{\mathsf{x}}(a-e)_{\mathsf{x}}} \end{align}