micro optimization - Intel 8086 Assembly -- Squaring a Register -

In principle, it is not hard to repay the value of a register:

  mov ax,  
 
 [Var] mov cx, [var] mul cx // class north dx: in ax  

But I got an opportunity to think - this course which I am learning for the award Efficiency is very high;

I know this is a micro-optimization, but will the following code work in the same way?

  mov ax, [var] mul ax / / is still the answer DX: Ax?  

I think that is a very easy way to express my question: AX (or AL / AH) mul and imul Order in the 8086 assembly?

Yes, mul ax puts ax = ax to dx: ax .