This question comes from QM but is cast as a purely mathematical question as follows:
Suppose operators $\sigma_i$ satisfy $[\sigma_i,\sigma_j] = 2i\epsilon_{ijk}\sigma_k$ and (n . $\sigma)^2 = 1$ for any unit vector n.
Show that (a . $\sigma$) (b . $\sigma$) = a . b + i a x b . $\sigma$ for any vectors a and b.
Contracting the commutation relation with $a_i$ and $b_j$ gives some progress but I somehow cannot make use of the other condition in any meaningful way. The only way I had 'managed' to incorporate the second condition was by saying that anti-commutator {$\sigma_i$,$\sigma_j$} had to be a multiple of $\delta_{ij}$ because it is symmetric, and then contracting with $n_i$ to get the constant of proportionality. But the starting argument is false as need isotropy rather than symmetry.
So after two more days, I found the solution to my problem;
Consider the symmetric anti-commutator $S_{ij}$ := {${\sigma_i, \sigma_j}$}. Contracting with $n_i$, $n_j$ and using the given relation (n . $\sigma)^2 = 1$ gives $n_iS_{ij}n_j =2$ $(*)$
Choose appropriate unit vectors n in $(*)$ to deduce that:
{${\sigma_i, \sigma_j}$} = $2 \delta_{ij} \ $
Adding equation $[\sigma_i,\sigma_j] = 2i\epsilon_{ijk}\sigma_k$ (given) to the above yields:
${\sigma_i \sigma_j} = \delta_{ij}+i\epsilon_{ijk}\sigma_k$.
Contracting above equation with $a_i$ and $b_j$ gives the required answer.