Squared scalar triple product from cyclic cross products ($(a\times b)\cdot [(b\times c)\times (c\times a)]=[a\cdot (b\times c){]}^2$)

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.