<s>[INST] Decomposition of a sum of cubes I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing  the squares. Can you give me a hint?
 [/INST] Your function is odd in $x$ and even in $y$, to wit
$[(-x)+2y]^3+[(-x)-2y]^3=-[(x-2y)^3+(x+2y)^3]$
$[x+2(-y)]^3+[x-2(-y)]^3=[(x-2y)^3+(x+2y)^3]$
So after factoring out $x$ your remaining quadratic factor must be even in both $x$ and $y$.  Therefore your red expression can only contain three nonzero terms $ax^2+by^2+c$.  Putting $x=0$ then forces $b=12,c=0$ and $y=0$ forces $a=1$.
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