mathematical puzzles

What are some interesting solutions to 33×33, if 11×11=4 and 22×22=16 ?

11∗11=4$11\ast 11=4$

But the 11 would be an addition of 1 and 1 instead. So:

(1+1)∗(1+1)=4$(1+1)\ast (1+1)=4$

2∗2=4$2\ast 2=4$

4=4$4=4$

The same rule would apply to 22 * 22 = 16, where the 22 becomes the addition of 2 and 2:

(2+2)∗(2+2)=16$(2+2)\ast (2+2)=16$

4∗4=16$4\ast 4=16$

16=16$16=16$

So now we apply the rule to 33*33:

33∗33=?$33\ast 33=?$

(3+3)∗(3+3)=?$(3+3)\ast (3+3)=?$

6∗6=36$6\ast 6=36$

?=36$?=36$

Hence the answer will be 36.