Think about this one:
| a | = | b |
| a2 | = | ab |
| a2-b2 | = | ab-b2 |
| (a+b)(a-b) | = | b(a-b) |
| (a+b) | = | b |
| a+a | = | a |
| 2a | = | a |
| 2 | = | 1 |
Once you understand the trick, you can "prove" anything you want, like 0=1 or 3=5.
Think about this one:
| a | = | b |
| a2 | = | ab |
| a2-b2 | = | ab-b2 |
| (a+b)(a-b) | = | b(a-b) |
| (a+b) | = | b |
| a+a | = | a |
| 2a | = | a |
| 2 | = | 1 |
Once you understand the trick, you can "prove" anything you want, like 0=1 or 3=5.
We start off with the axiom that a = b.
At step three you have zero on each side because a^2-b^2 = 0, and ab-b^2 = 0.
Sure you can still keep doing algebra after this point but, as your result shows, all it does is lead to an apparent contradiction. The contradiction does not actually exist. It's just bad math.
I think the biggest flaw is going from step 4 to step 5, where you divide by (a-b), and since a-b=0, you are dividing by zero which is undefined. After that point it's bad math.