= Solution
Expand the same double <cross product> in two ways. First expanding with $c\times d$ as the inner <cross product> gives
$$
(a\times b)\times(c\times d)
=[(a\times b)\cdot d]c-[(a\times b)\cdot c]d
=[a,b,d]c-[a,b,c]d.
$$
Alternatively, use $(a\times b)\times w=(a\cdot w)b-(b\cdot w)a$ with $w=c\times d$ to get
$$
(a\times b)\times(c\times d)=[a,c,d]b-[b,c,d]a.
$$
Equating these expressions and moving every term to one side proves
$$
[b,c,d]a-[a,c,d]b+[a,b,d]c-[a,b,c]d=0.
$$
This <four-vector scalar triple product identity> is valid without assuming that any three of the four <vectors> are independent.
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