Theorem 1.
If $\mathbf{a},\mathbf{b},\mathbf{c}\in\mathbf{R}^3$ then $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$ can be written in the following simpler way
\begin{equation}
\label{eq:tripleproduct}
\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\langle\mathbf{a},\mathbf{c}\rangle\mathbf{b}-
\langle\mathbf{a},\mathbf{b}\rangle\mathbf{c}.
\end{equation}
Lemma.
\begin{equation}\label{eq:scalartripleprod}
[\mathbf{x}\mathbf{y}\mathbf{z}]:=\langle\mathbf{x}, \mathbf{y}\times\mathbf{z}\rangle=\langle\mathbf{x}\times\mathbf{y}, \mathbf{z}\rangle,
\end{equation}
and
\begin{equation}
\label{eq:specscalartripleprod}
\mathbf{n}\times(\mathbf{c}\times\mathbf{n})=\|\mathbf{n}\|^2\mathbf{c}\quad\text{for}\quad\mathbf{n}\perp\mathbf{c}.
\end{equation}
If $\mathbf{b}=\mathbf{0}$ or $\mathbf{c}=\mathbf{0}$ then both sides of \eqref{eq:tripleproduct} are $\mathbf{0}$.
If $\mathbf{b}\parallel\mathbf{c}$ then both sides of \eqref{eq:tripleproduct} are $\mathbf{0}$.
Now assume that $\mathbf{b}$ and $\mathbf{c}$ are linearly independent.
The perpendicular property of the cross-product implies that
\begin{equation}
\label{eq:alphabeta}
\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\alpha\mathbf{b}+\beta\mathbf{c}.
\end{equation}
Let $\mathbf{n}:=\mathbf{b}\times\mathbf{c}$. Taking the dot product of both sides of \eqref{eq:alphabeta} with $\mathbf{c}\times\mathbf{n}/[\mathbf{b}\mathbf{c}\mathbf{n}]$ and using that $[\mathbf{b}\mathbf{c}\mathbf{n}]=\langle\mathbf{b}\times\mathbf{c},\mathbf{n}\rangle=\langle\mathbf{n},\mathbf{n}\rangle=\|\mathbf{n}\|^2$ we obtain
\[
\begin{align*}
\left\langle\mathbf{a}\times(\mathbf{b}\times\mathbf{c}), \frac{\mathbf{c}\times\mathbf{n}}{[\mathbf{b}\mathbf{c}\mathbf{n}]}\right\rangle &=\alpha\frac{\langle\mathbf{b}, \mathbf{c}\times\mathbf{n}\rangle}{[\mathbf{b}\mathbf{c}\mathbf{n}]} +\beta \frac{\langle\mathbf{c}, \mathbf{c}\times\mathbf{n}\rangle}{[\mathbf{b}\mathbf{c}\mathbf{n}]} \\
\frac{\langle\mathbf{a}\times\mathbf{n},\mathbf{c}\times\mathbf{n} \rangle}{\|\mathbf{n}\|^2}&=\alpha\cdot 1+\beta\cdot 0\\
\frac{\langle\mathbf{a}, \mathbf{n}\times(\mathbf{c}\times\mathbf{n})\rangle}{\|\mathbf{n}\|^2}&=\alpha\\
\frac{\langle\mathbf{a},\|\mathbf{n}\|^2\mathbf{c}\rangle}{\|\mathbf{n}\|^2}&=\alpha \quad\text{by \eqref{eq:specscalartripleprod}}\\
\langle\mathbf{a},\mathbf{c}\rangle&=\alpha.
\end{align*}
\]
We have
\[
\mathbf{a}\times(\mathbf{c}\times\mathbf{b})=(-\beta)\mathbf{c}+(-\alpha)\mathbf{b}.
\]
Now starting from \eqref{eq:alphabeta} interchanging $\mathbf{b}$ and $\mathbf{c}$ we get
\[
\langle\mathbf{a},\mathbf{b}\rangle=-\beta,
\]
that is,
\[
-\langle\mathbf{a},\mathbf{b}\rangle=\beta,
\]
which yields \eqref{eq:tripleproduct}. $\blacksquare$
If $\mathbf{b}\parallel\mathbf{c}$ then both sides of \eqref{eq:tripleproduct} are $\mathbf{0}$.
Now assume that $\mathbf{b}$ and $\mathbf{c}$ are linearly independent.
The perpendicular property of the cross-product implies that
\begin{equation}
\label{eq:alphabeta}
\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\alpha\mathbf{b}+\beta\mathbf{c}.
\end{equation}
Let $\mathbf{n}:=\mathbf{b}\times\mathbf{c}$. Taking the dot product of both sides of \eqref{eq:alphabeta} with $\mathbf{c}\times\mathbf{n}/[\mathbf{b}\mathbf{c}\mathbf{n}]$ and using that $[\mathbf{b}\mathbf{c}\mathbf{n}]=\langle\mathbf{b}\times\mathbf{c},\mathbf{n}\rangle=\langle\mathbf{n},\mathbf{n}\rangle=\|\mathbf{n}\|^2$ we obtain
\[
\begin{align*}
\left\langle\mathbf{a}\times(\mathbf{b}\times\mathbf{c}), \frac{\mathbf{c}\times\mathbf{n}}{[\mathbf{b}\mathbf{c}\mathbf{n}]}\right\rangle &=\alpha\frac{\langle\mathbf{b}, \mathbf{c}\times\mathbf{n}\rangle}{[\mathbf{b}\mathbf{c}\mathbf{n}]} +\beta \frac{\langle\mathbf{c}, \mathbf{c}\times\mathbf{n}\rangle}{[\mathbf{b}\mathbf{c}\mathbf{n}]} \\
\frac{\langle\mathbf{a}\times\mathbf{n},\mathbf{c}\times\mathbf{n} \rangle}{\|\mathbf{n}\|^2}&=\alpha\cdot 1+\beta\cdot 0\\
\frac{\langle\mathbf{a}, \mathbf{n}\times(\mathbf{c}\times\mathbf{n})\rangle}{\|\mathbf{n}\|^2}&=\alpha\\
\frac{\langle\mathbf{a},\|\mathbf{n}\|^2\mathbf{c}\rangle}{\|\mathbf{n}\|^2}&=\alpha \quad\text{by \eqref{eq:specscalartripleprod}}\\
\langle\mathbf{a},\mathbf{c}\rangle&=\alpha.
\end{align*}
\]
We have
\[
\mathbf{a}\times(\mathbf{c}\times\mathbf{b})=(-\beta)\mathbf{c}+(-\alpha)\mathbf{b}.
\]
Now starting from \eqref{eq:alphabeta} interchanging $\mathbf{b}$ and $\mathbf{c}$ we get
\[
\langle\mathbf{a},\mathbf{b}\rangle=-\beta,
\]
that is,
\[
-\langle\mathbf{a},\mathbf{b}\rangle=\beta,
\]
which yields \eqref{eq:tripleproduct}. $\blacksquare$
Theorem 2.
If $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\in\mathbf{R}^3$ then
\begin{equation}
\label{eq:quadrupleproduct}
(\mathbf{a}\times\mathbf{b})\times(\mathbf{c}\times\mathbf{d})=[\mathbf{a}\mathbf{b}\mathbf{d}]\mathbf{c}-
[\mathbf{a}\mathbf{b}\mathbf{c}]\mathbf{d}.
\end{equation}
From \eqref{eq:tripleproduct} it follows
\begin{align*}
(\mathbf{a}\times\mathbf{b})\times(\mathbf{c}\times\mathbf{d}) &=\langle\mathbf{a}\times\mathbf{b},\mathbf{d}\rangle\mathbf{c}-
\langle\mathbf{a}\times\mathbf{b},\mathbf{c}\rangle\mathbf{d}\\
&=[\mathbf{a}\mathbf{b}\mathbf{d}]\mathbf{c}-
[\mathbf{a}\mathbf{b}\mathbf{c}]\mathbf{d}.\qquad\blacksquare \end{align*}
\begin{align*}
(\mathbf{a}\times\mathbf{b})\times(\mathbf{c}\times\mathbf{d}) &=\langle\mathbf{a}\times\mathbf{b},\mathbf{d}\rangle\mathbf{c}-
\langle\mathbf{a}\times\mathbf{b},\mathbf{c}\rangle\mathbf{d}\\
&=[\mathbf{a}\mathbf{b}\mathbf{d}]\mathbf{c}-
[\mathbf{a}\mathbf{b}\mathbf{c}]\mathbf{d}.\qquad\blacksquare \end{align*}
Corollary.
If $\mathbf{a},\mathbf{b},\mathbf{c}\in\mathbf{R}^3$ then\[
\langle\mathbf{a},\mathbf{b}\times\mathbf{c}\rangle\mathbf{a}=(\mathbf{a}\times\mathbf{b})
\times(\mathbf{a}\times\mathbf{c}). \]
Theorem 3.
If $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\in\mathbf{R}^3$ then
\[
\langle\mathbf{a}\times\mathbf{b},\mathbf{c}\times\mathbf{d}\rangle=\langle\mathbf{a},\mathbf{c}\rangle\langle\mathbf{b},\mathbf{d}\rangle-\langle\mathbf{a},\mathbf{d}\rangle\langle\mathbf{b},\mathbf{c}\rangle.
\]
\begin{align*}
\langle\mathbf{a}\times\mathbf{b},\mathbf{c}\times\mathbf{d}\rangle &=[\mathbf{a}\mathbf{b}(\mathbf{c}\times\mathbf{d})] \\
&=\langle\mathbf{a},\mathbf{b}\times(\mathbf{c}\times\mathbf{d})\rangle\quad\text{by \eqref{eq:scalartripleprod}}\\
&=\langle\mathbf{a},\langle\mathbf{b},\mathbf{d}\rangle\mathbf{c}-\langle\mathbf{b},\mathbf{c}
\rangle\mathbf{d}\rangle\quad\text{by \eqref{eq:tripleproduct}}\\
&=\langle\mathbf{a},\mathbf{c}\rangle\langle\mathbf{b},\mathbf{d}\rangle-\langle\mathbf{a},\mathbf{d}\rangle\langle\mathbf{b},\mathbf{c}\rangle.\qquad\blacksquare \end{align*}
\langle\mathbf{a}\times\mathbf{b},\mathbf{c}\times\mathbf{d}\rangle &=[\mathbf{a}\mathbf{b}(\mathbf{c}\times\mathbf{d})] \\
&=\langle\mathbf{a},\mathbf{b}\times(\mathbf{c}\times\mathbf{d})\rangle\quad\text{by \eqref{eq:scalartripleprod}}\\
&=\langle\mathbf{a},\langle\mathbf{b},\mathbf{d}\rangle\mathbf{c}-\langle\mathbf{b},\mathbf{c}
\rangle\mathbf{d}\rangle\quad\text{by \eqref{eq:tripleproduct}}\\
&=\langle\mathbf{a},\mathbf{c}\rangle\langle\mathbf{b},\mathbf{d}\rangle-\langle\mathbf{a},\mathbf{d}\rangle\langle\mathbf{b},\mathbf{c}\rangle.\qquad\blacksquare \end{align*}
