Proof in scalar triple product identities How can I prove it? I could not do it by using determinant.
Let $u,v,w,t$ $\in$ $\Bbb R^3$. Show that quaternary identities below are equal
$[u,v,w]t = [t,v,w]u + [u,t,w]v + [u,v,t]w$
 A: Note that proving this identity is equivalent to proving the following identities for each component of the vectors with $i\in\{1,2,3\}$:
$$[u,v,w]t_i = [t,v,w]u_i + [u,t,w]v_i + [u,v,t]w_i$$
Therefore, construct the following determinant 
$$\left|\begin{array}{cccc}u_i & u_1 & u_2 & u_3 \\ v_i & v_1 & v_2 & v_3 \\ w_i & w_1 & w_2 & w_3 \\ t_i & t_1 & t_2 & t_3 \end{array}\right|$$
It is clear that this determinant is zero because two columns are identical, depending on which value $i$ takes, it will be the first column and the $i+1$th.
Now, develop this determinant with respect to the first column
$$\left|\begin{array}{ccc} v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \\ t_1 & t_2 & t_3 \end{array}\right|u_i - \left|\begin{array}{ccc} u_1 & u_2 & u_3 \\ w_1 & w_2 & w_3 \\ t_1 & t_2 & t_3 \end{array}\right|v_i + \left|\begin{array}{ccc} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ t_1 & t_2 & t_3 \end{array}\right|w_i-\left|\begin{array}{ccc} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{array}\right|t_i$$
or in the triple product notation
$$[v,w,t]u_i - [u,w,t]v_i + [u,v,t]w_i - [u,v,w]t_i $$
For the first term, two permutations transform the triple product in $[t,v,w]$ leaving the sign unchanged. For the second term, one permutation transforms it to $[u,t,w]$, changing the sign. Thus we have
$$[t,v,w]u_i + [u,t,w]v_i + [u,v,t]w_i - [u,v,w]t_i $$
And since this is equal to zero, we have your identity.
