Factorise of the expression $$ a^4+b^4-c^4-2a^2b^2+4abc^2~ \\(S:(a + b - c) (a + b + c) (a^2+ b^2 + c^2 - 2 a b))$$
I tried to solve this problem: $$ a^4+b^4-c^4-2a^2b^2+4abc^2\rightarrow\\\ (a^2-b^2)^2-c^4+4abc^2\rightarrow\\\ (a^2-b^2-c^2)(a^2-b^2+c^2)+4abc^2\\\ ((a-b)(a+b)-c^2)((a-b)(a+b)+c^2)+4abc^2$$ but I stopped here. Can anyone help me?
On
As a quadratic in $c^2\,$, the reduced discriminant of $-(c^4-4ab \cdot c^2-a^4-b^4+2a^2b^2)$ is:
$$ \frac{1}{4}\Delta = 4a^2b^2 +a^4+b^4-2a^2b^2=(a^2+b^2)^2 $$
Therefore the roots are $c^2=2ab \pm (a^2+b^2) = \{(a+b)^2, -(a-b)^2\}\,$ giving the factorization:
$$ \begin{align} -\left(c^2-(a+b)^2\right)\left(c^2+(a-b)^2\right) & = -(c-a-b)(c+a+b)(c^2+a^2-2ab+b^2) \\[3px] & = (a+b-c)(a+b+c)(a^2+b^2+c^2-2ab) \end{align} $$
.You should actually do: \begin{split} a^4+b^4-c^4-2a^2b^2+4abc^2 & = (a^4+b^4 + 2a^2b^2) - (4a^2b^2 + c^4 - 4abc^2) \\ & = (a^2+b^2)^2 - (c^2-2ab)^2 \\ & = (a^2+b^2-c^2+2ab)(a^2+b^2+c^2-2ab) \\ & = ((a+b)^2 - c^2) (a^2+b^2+c^2-2ab) \\ & = (a+b+c)(a+b-c)(a^2+b^2+c^2-2ab) \end{split}
In the first step, I added and subtracted $4a^2b^2$, and in the second step grouped them into squares. From then, it is just the formula $x^2-y^2 = (x+y)(x-y)$ applied twice.
In your attempt, the $4abc^2$ was just left hanging. That shoul have been dealt with by trying to bring it into the expansion of a square term (which I did).