Factor $ a^4-3a^2-2ab+1-b^2$ In order to factor $a^4-3a^2-2ab+1-b^2$, I find that $a=1, b=-1$ makes the value of the expression 0. Thus, I assume $b=-a$.
I rewrite the expression on the assumption as:
$$a^4-3a^2-2ab+1-b^2$$
$$=a^4-3a^2+2a^2+1-a^2$$
$$=a^4-2a^2+1$$
$$=(a^2-1)^2$$
Then I insert $a+b$, which is another form of the assumption above, into one of the factors.  Since $a+b=0$, this insertion should cause no change to whatever relationships in the original expression.  I arbitrarily set its coefficient as $1$.  A factor, therefore, would be $(a^2+a+b-1)$.
I then divide the original expression by the factor and find that it indeed can be divided without a remainder and I also find the other factor.  Thus,
$$a^4-3a^2-2ab+1-b^2$$
$$=(a^2+a+b-1)(a^2-a-b-1)$$
Q.E.D.
The problem is that I don't know what I am doing.  I tried to use the factor theorem for a bivariate (if this is the proper word for it) expression but I might be fabricating a fortuitous reasoning from the answer.  Yes, I knew the answer before I tried, and no, I don't know how I can solve this in other ways.  It would be great if someone can tell me if my reasoning can be made more clear.

Here is probably the most simple method to factor it.
Note that $$a^2-b^2=(a-b)(a+b)$$So $$a^4-3a^2-2ab+1-b^2=a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$$
So we can factor it into $$(a^2+a+b-1)(a^2-a-b-1)$$
