Categorical product cancellable

Question

It is pretty easy to prove that $A\cong B\Rightarrow A\times X\cong B\times X$ (assuming the products exist). However, I haven't been able to prove the other direction, that is, $A\times X\cong B\times X \Rightarrow A\cong B$ (cancellation of an equal factor, therefore the title). My attempt:
Because of isomorphism, there have to be arrows $t_A: B\times X\rightarrow A\times X$ and $t_B: A\times X\rightarrow B\times X$ with $t_A\circ t_B=id_{A\times X}$, $t_B\circ t_A=id_{B\times X}$, as well as (by definition of products) $f_A: A\times X\rightarrow A$, $f_X: A\times X\rightarrow X$, $g_B: B\times X\rightarrow B$, $g_X: B\times X\rightarrow X$ satisfying the universal property of $A\times X$ and $B\times X$. From these we get arrows $f_B: A\times X\rightarrow B=g_B\circ t_B$, $g_A: B\times X\rightarrow A=f_A\circ t_A$.
However, now I got stuck, because there are no morphisms from $A$ or $B$ (that I see). So my questions are:
1. Is $A\cong B\Leftrightarrow A\times X\cong B\times X$?
2. If it is, how would you prove it?
3. If not, what additional properties would be needed to prove it?