<s>[INST] 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.
 [/INST] 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 $$(a^2+a+b-1)(a^2-a-b-1)$$
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