Given a 3-braid $b=\sigma_1\sigma_2^{-1}\sigma_1\sigma_2^{-1}\sigma_1$ (which has non-trivial closure), can we find a 3-braid $c$, which has trivial closure (closure results in any trivial knot or link), such that $cb$ has a trivial closure?
My thinking so far: From this answer(and the comments), we know that a 3-braid with trivial closure must be in the form $gag^{-1}$, where $a\in\{\sigma_1\sigma_2,\sigma_1^{-1}\sigma_2^{-1},\sigma_1\sigma_2^{-1},\sigma_1,e\}$, $g$ is any 3-braid. $e$ denotes identity.
So we have $c=g_1a_1g_1^{-1}$, $cb=g_2a_2g_2^{-1}$, hence $g_1a_1g_1^{-1}b=g_2a_2g_2^{-1}$. I am not sure if this is useful, or we need something else to show it.
