We first prove that
\begin{align*}
\mathbf{c} \times (\mathbf{b} \times \mathbf{a}) \cdot \mathbf{d} = (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}).
\end{align*}
This can be shown by straightforward matrix algebra using the correspondence between elements of \(\mathbb{R}^3\) and \(\mathfrak{so}(3)\), given by
\(\mathbb{R}^3 \ni \mathbf{a} = \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}^\mathrm{T} \mapsto \mathbf{\hat{a}} \in \mathfrak{so}(3)\), where
\begin{align*}
\mathbf{\hat{a}} = \begin{bmatrix}
0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0
\end{bmatrix}.
\end{align*}
It then follows from the properties of
skew-symmetric matrices that
\begin{align*}
\mathbf{c} \times (\mathbf{b} \times \mathbf{a}) \cdot \mathbf{d} = (\mathbf{\hat{c}}\mathbf{\hat{b}}\mathbf{a})^\mathrm{T} \mathbf{d} = \mathbf{a}^\mathrm{T} \mathbf{\hat{b}} \mathbf{\hat{c}}\mathbf{d} = (-\mathbf{\hat{b}}\mathbf{a})^\mathrm{T} \mathbf{\hat{c}}\mathbf{d} = (\mathbf{\hat{a}}\mathbf{b})^\mathrm{T}\mathbf{\hat{c}}\mathbf{d} = (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}).
\end{align*}
We also know from
vector triple products that
\begin{align*}
\mathbf{c} \times (\mathbf{b} \times \mathbf{a}) = (\mathbf{c} \cdot \mathbf{a})\mathbf{b} - (\mathbf{c} \cdot \mathbf{b})\mathbf{a}.
\end{align*}
Using this identity along with the one we have just derived, we obtain the desired identity:
\begin{align*}
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = \mathbf{c} \times (\mathbf{b} \times \mathbf{a}) \cdot \mathbf{d} = \left[ (\mathbf{c} \cdot \mathbf{a})\mathbf{b} - (\mathbf{c} \cdot \mathbf{b})\mathbf{a} \right] \cdot \mathbf{d} = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c}).
\end{align*}
