= Squared scalar triple product from cyclic cross products
{title2=$(a\times b)\cdot[(b\times c)\times(c\times a)]=[a\cdot(b\times c)]^2$}
Apply the <vector triple product identity> with first vector $b\times c$. Its contraction with $c$ vanishes, so $(b\times c)\times(c\times a)=[a\cdot(b\times c)]c$. Dotting with $a\times b$ gives the square of the <scalar triple product>. This is also the <Gram determinant> of $a,b,c$, hence is nonnegative even when the vectors are linearly dependent.
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