Factorise: $a^4-b^4+c^4-d^4-2(a^2c^2-b^2d^2)+4ac(b^2+d^2)-4bd(a^2+c^2)$

Question

Factorise: $a^4-b^4+c^4-d^4-2(a^2c^2-b^2d^2)+4ac(b^2+d^2)-4bd(a^2+c^2)$.

My working: $(a^4-2a^2c^2+c^4)-(b^4-2b^2d^2+d^4)+4ac(b^2+d^2)-4bd(a^2+c^2)$
$=(a^2-c^2)^2-(b^2-d^2)^2+4ac(b^2+d^2)-4bd(a^2+c^2)$
$=(a+c)^2(a-c)^2-(b+d)^2(b-d)^2+4ac(b^2+d^2)-4bd(a^2+c^2)$

Solution: $(a+c+b+d)(a+c-b-d)(a^2-2ac+c^2+b^2-2bd+d^2)$

My question is: it is not hard to work backward from the solution and see how it works, but how could I proceed from my stage of working if we were not told about the solution? And how many ways are there to factorise this expression?

Many thanks!

Answer 1

Last question first: If you take integers coefficients and factor all the way down to irreducibles (no further factoring possible), then up to ordering, there is only one factorization.

Factoring this one by hand is all about recognizing differences of squares. The hints are "$a^4+c^4$" with "$a^2+c^2$" and "$a^2c^2$" and likewise for $b$ with $d$. The general method is "you don't want to know". If there's interest, I can describe an entirely automatable method that no human should ever undertake.