Question: finding cube of a number

[Visual Basic.Net]

I wrote this to multiply the x by itself 3times then add a number y to it and save result in z. but it dosn't giving me the right result.

.DATA
X DB 12h
Y DB 34h
Z DW ?,?

mov al,x
mov bl,x
mul bl; ax=bl*al
mul ax; dxax=ax*ax
add ax,y
mov z,ax
mov z+2,bx

I wrote it in Debug.
thank's

[Pradeep1210]

Is that .NET or Assembly.. or IL ??

[Visual Basic.Net]

It is assembly code dealing with 8086 processor.
I want to find the cube of a number.
My problem, for ex:
MOV AX,0012h;
MUL AX; that will do DXAX=AX*AX

now how can I multiply the DXAX with AX again to get the cube stored in DXAX ?

thank's in advance

[minor28]

If x always is 2 i.e 2^3 you can do it like this

mov eax,2
shl eax,2

Result in eax

If you use mul you do it like this (5^3)

mov eax,5
mov ecx,2
jmp @F
back:
mov edx,5
mul edx
dec ecx
@@:
cmp ecx,0
ja back

Result in eax

If you use co-processor you do i like this (5^3)

finit	;Initialize the co-processor
mov eax,0		;Load
push eax		;3
mov eax,3		;to stack
push eax		;as a qword
fild qword ptr [esp]	;Load 3 from stack to 80-bit register st(0)=3
add esp,8		;Restore stack
mov eax,0		;Load
push eax		;5
mov eax,5		;to stack
push eax		;as a qword
fild qword ptr [esp]	;Load 5 from stack to next register  st(0)=5, st(1)=3
add esp,8		;Restore stack
fyl2x			;st(0)=st(1)*log2(st(0)) => st(0)=5*log2(3)
fst st(1)		;Store st(1)=st(0)=5*log2(3)
frndint			;Round ST(0) to an integer
fsub st(1),st(0)	;Subtract st(0) from st(1) => st(0)=integer and st(1)=fraction
fxch st(1)		;Exchange st(0) and st(1)
f2xm1			;computes the exponential value of 2 to the power of the value of ST(0) and subtracts 1
fld1			;Load value 1 to st(0). st(0)=1, st(1)=2^st(0), st(2)=integer
fadd			;st(0)=2^st(0), st(1)=integer
fscale			;Scale ST(0) by ST(1)
fwait			;Wait while co-processor is busy
fstp fResult		;Store in fResult(qword) and pop ST(0)