Factorize $\displaystyle a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)$
It seems like some obvious combination like $\displaystyle (\pm a^2\pm b^2\pm c)^2$ doesn't work for any combination of +/-
We can write it as a polynomial g with:
$\displaystyle c=x$
and
$\displaystyle g(x)= x^2-2(a^2+b^2)x+(a^4+b^4-2a^2b^2)$
and then use abc-formula to find roots of $\displaystyle g(x)=0$
Then we can write $\displaystyle (x-r_1)(x-r_2)$ as a factorisation. (r_1,r_2 roots)
But such factorisation is not quite unique, in this case...
Dear vuze88,
Notice that, $\displaystyle (a^2+b^2-c)^2=a^4+b^4+c^2+2a^2b^2-2a^2c-2b^2c$
Therefore, $\displaystyle (a^2+b^2-c)^2-4a^2b^2=a^4+b^4+c^2-2a^2b^2-2a^2c-2b^2c$
$\displaystyle a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)=(a^2+b^2-c)^2-4a^2b^2=(a^2+b^2-c-2ab)(a^2+b^2-c+2ab)$
$\displaystyle a^4+b^4+c^2-2(a^2b^2+a^2c+b^2c)=((a-b)^2-c)((a+b)^2-c)$
Hope this helps.