How can I use Fundamental Theorem of Symmetric Polynomials to factor polynomials? 
How can I use The fundamental theorem of symmetric polynomials (or its proof) to factor symmetric polynomials?  

The link I've given to the theorem uses elaborate wordings using 'rings', 'isomorphic', etc.  
I completely understand those objects or describings are needed to have a deep understanding, but could anyone try, if it is possible, to explain simply how I could use the theorem to, e.g., factor
$(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2 = a^2b^2(a^2+b^2-2ab)$  
without understanding what rings are? I only wish to be able to practically use it.
 A: If I understand correctly, the fundamental theorem of symmetric polynomials says a symmetric polynomial of $x_1, x_2, \dots, x_n$ can be written as a polynomial of $$\quad e_1 = \sum_i x_i,\quad e_2 = \sum_{i < j}x_ix_j, \quad \dots, \quad e_n = \prod_i x_i.$$
For example, symmetric polynomial $(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2$ can be written as a polynomial of $x = a + b, y = ab$. In this example, inductively one can get $a^n + b^n$ in terms of $x,y$. Note that $$a^2 + b^2 = (a+b)^2-2ab = x^2-2y,$$ $$a^3 + b^3 = (a^2 + b^2)(a+b) - ab(a+b) = (x^2-2y)x - yx = x^3-3xy,$$ $$a^4+b^4 = (a^3+b^3)(a+b)-ab(a^2+b^2) = (x^3-3xy)x-y(x^2-2y)=x^4-4x^2y+2y^2.$$
Therefore we have $$(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2 = (x^2-2y)(x^4-4x^2y+2y^2)-(x^3-3xy)^2 = x^2y^2-4y^3.$$
In general, once a symmetric polynomial is written in terms of elementary symmetric polynomials, it might be easier to observe factors. In this example, $x^2y^2-4y^3 = y^2(x^2-4y)$. 
Warning One might want to substitute the original variables back to see if certain factors can be factorized further. In the example, $x^2-4y$ is irreducible, however $x^2-4y = a^2+b^2-2ab = (a-b)^2$ is reducible.
A: I will give you another algoritm to find the factorization. I'll use your polynomial to show the way it works. For a polynomial in $k$ variables the algorithm can be described as:


*

*Order the terms of your polynomial in lexicographic ordening.

*Suppose the highest term is $X_{1}^{n_1}\cdots X_{k}^{n_k}$, then subtract $e_{1}^{n_1-n_2}e_{2}^{n_2-n_3}\cdots e_{k}^{n_k}$.

*Remember this term and head back to step 1 with your new polynomial. If the new polynomial is zero, you are finished.

*Add all the terms found to get your representation for the original polynomial.


I won't prove here that it works, I will just show you how to apply it. The lexicographic ordening means that we start with the term which has highest $n_1$, then if two have the same value for $n_1$ we choose the one with highest $n_2$ etc. 
Your polynomial: $f = (a^4+b^4)(a^2+b^2)−(a^3+b^3)^2 = a^4b^2 -2a^3b^3+ a^2b^4$,  put in the lexicographic ordening. Next we create 
\begin{equation}f_1 = f - e_{1}^{4-2}e_{2}^{2} = f-(a+b)^2(ab)^2 = (a^4b^2-2a^3b^3+ a^2b^4) - (a^4b^2+2a^3b^3 + a^2b^4) = 0.
\end{equation}
Since this is zero we are done after one step and conclude that $f = f_1 + e_{1}^{4-2}e_{2}^{2} = 0 + e_{1}^{4-2}e_{2}^{2} = (a+b)^2(ab)^2$.
A: I just wanna add an easy way for readers: (it can be used in Projective Spaces in Algebraic Geometry)
First put $b=1$ (this is dehomogenize). Then $$(a^4+1)(a^2+1)-(a^3+1)^2=a^2(a^2+1-2a).$$ Now homoginize it to degree $6$: $$a^2b^2(a^2+b^2-2ab).$$
