How Seriously... (2 Viewers)

[Affinity]

a^4 + (a^2)(b^2) + b^4 = a^4 + 2(a^2)(b^2) + b^4 -(a^2)(b^2)
= a^4 + 2(ab)^2 + b^4 -(ab)^2
= (a^2 + b^2)^2 -(ab)^2
= (a^2 + b^2 + ab)(a^2 + b^2 -ab)

I find it more natural to see it this way:

a^4 + (a^2)(b^2) + b^4 = (a^6 - b^6)/(a^2 - b^2)
= (a^3 - b^3)(a^3 + b^3)/(a+b)(a-b)
= (a-b)(a^2 + ab + b^2)(a+b)(a^2 - ab + b^2)/(a+b)(a-b)
= answer.

it's interesting too in that one expands the set of objects under consideration for the purpose of calculation.

 

[Slidey]

I find it more natural to see it this way:

a^4 + (a^2)(b^2) + b^4 = (a^6 - b^6)/(a^2 - b^2)
= (a^3 - b^3)(a^3 + b^3)/(a+b)(a-b)
= (a-b)(a^2 + ab + b^2)(a+b)(a^2 - ab + b^2)/(a+b)(a-b)
= answer.

it's interesting too in that one expands the set of objects under consideration for the purpose of calculation.

I prefer this method:
a^4 + (a^2)(b^2) + b^4
(a^2 + b^2 + b^4/a^2)*a^2

z=a+b^2/a
z^2=a^2+2b^2+b^4/a^2, so we get:

(z^2-b^2)*a^2
(z+b)(z-b)*a^2
(a+b^2/a+b)(a+b^2/a-b)*a^2
(a^2+b^2+ab)(a^2+b^2-ab)

Of course, it's actually a general method for quasi-symmetric equation, which this (and all sum of two squares methods for quartics) is a trivial case of.