How did the author arrive at step $(3)$ from step $(1)$ and step $(2)$ in the following definition of vector quadruple product?
Many combinations of vector and scalar products are possible, but we consider only one more, namely the vector quadruple product $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})$. By regarding $\mathbf{a} \times \mathbf{b}$ as a single vector, we see that this vector must be representable as a linear combination of $\mathbf{c}$ and $\mathbf{d}$. On the other hand, regarding $\mathbf{c} \times \mathbf{d}$ as a single vector, we see that it must also be a linear combination of $\mathbf{a}$ and $\mathbf{b}$. This provides a means of expressing one of the vectors, say $\mathbf{d}$, as linear combination of the other three, as follows:
\begin{align}
(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d}) &=[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}] \mathbf{c}-[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}] \mathbf{d} \tag{1}\\
&=[(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{a}] \mathbf{b}-[(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{b}] \mathbf{a} \tag{2}
\end{align}
Hence
$$
[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}] \mathbf{d}=[(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{d}] \mathbf{a}+[(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{d}] \mathbf{b}+[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}] \mathbf{c}\tag{3}
$$
or
$$
\mathbf{d}=\frac{[(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{d}] \mathbf{a}+[(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{d}] \mathbf{b}+[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}] \mathbf{c}}{[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}]}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}
$$
Best Answer
We have: $[(a\times b)\cdot d]c - [(a\times b)\cdot c]d = [(c\times d)\cdot a]b - [(c\times d)\cdot b]a$
isolating the intended left side we get $[(a\times b)\cdot c]d = [(c\times d)\cdot b]a - [(c\times d)\cdot a]b + [(a\times b)\cdot d]c$
Recognizing the things in the brackets as the scalar triple product, we can rearrange the first term by cycling $(c\times d) \cdot b = (b\times c) \cdot d$, and rearrange the second term and its sign with a single order swap $-(c\times d)\cdot a = (c \times a)\cdot d$. Plug these manipulations back in and we get the intended result, $$[(a\times b)\cdot c]d = [(b\times c)\cdot d]a + [(c\times a)\cdot d]b + [(a\times b)\cdot d]c$$