= Solution
Expand the <vector triple product> in Cartesian components. Its first component is
$$
\begin{aligned}
[\mathbf a\times(\mathbf b\times\mathbf c)]_x
&=a_y(b_xc_y-b_yc_x)-a_z(b_zc_x-b_xc_z)\\
&=b_x(a_yc_y+a_zc_z)-c_x(a_yb_y+a_zb_z)\\
&=b_x(\mathbf a\cdot\mathbf c)-c_x(\mathbf a\cdot\mathbf b),
\end{aligned}
$$
where adding $a_xb_xc_x$ to both products introduces cancelling terms. Likewise the other two components are
$$
\begin{aligned}
[\mathbf a\times(\mathbf b\times\mathbf c)]_y
&=a_z(b_yc_z-b_zc_y)-a_x(b_xc_y-b_yc_x)
=b_y(\mathbf a\cdot\mathbf c)-c_y(\mathbf a\cdot\mathbf b),\\
[\mathbf a\times(\mathbf b\times\mathbf c)]_z
&=a_x(b_zc_x-b_xc_z)-a_y(b_yc_z-b_zc_y)
=b_z(\mathbf a\cdot\mathbf c)-c_z(\mathbf a\cdot\mathbf b).
\end{aligned}
$$
Together these prove
$$
\boxed{\mathbf a\times(\mathbf b\times\mathbf c)
=(\mathbf a\cdot\mathbf c)\mathbf b-(\mathbf a\cdot\mathbf b)\mathbf c}.
$$
Now use the cyclic property of the <scalar triple product> and the just-proved identity:
$$
\begin{aligned}
(\mathbf a\times\mathbf b)\cdot(\mathbf c\times\mathbf d)
&=\mathbf a\cdot[\mathbf b\times(\mathbf c\times\mathbf d)]\\
&=\mathbf a\cdot[(\mathbf b\cdot\mathbf d)\mathbf c-(\mathbf b\cdot\mathbf c)\mathbf d].
\end{aligned}
$$
This proves <Lagrange's identity> in the requested form,
$$
\boxed{(\mathbf a\times\mathbf b)\cdot(\mathbf c\times\mathbf d)
=(\mathbf a\cdot\mathbf c)(\mathbf b\cdot\mathbf d)
-(\mathbf a\cdot\mathbf d)(\mathbf b\cdot\mathbf c)}.
$$
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