Decomposition of a sum of cubes

Question

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?

Answer 1

Hint: Use $a^2+b^2-ab=(a-b)^2+ab$

Answer 2

Compare $S = (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
with $T = (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.$

Adding: $S + T = 2(a^3 + 3ab^2).$

Answer 3

hint

put $$z=2y$$

and use

$$a^3+b^3=(a+b)(a^2-2ab+b^2+ab)$$ $$=(a+b)\Bigl((a-b)^2+ab\Bigr)$$

where $$a=x+z \text{ and } b=x-z$$

Answer 4

Use $(a+b)+(a-b)=2a$
and $(a+b)^2+(a-b)^2=2a^2+2b^2$
and $(a-b)(a+b)=a^2-b^2$
Then $$ \underbrace{(\color{#C00}{(x+2y)}+\color{#090}{(x-2y)})}_{2x}\left(\!\vphantom{2^2}\right.\underbrace{\color{#C00}{(x+2y)}^2+\color{#090}{(x-2y)}^2}_{2x^2+8y^2}-\underbrace{\color{#C00}{(x+2y)}\color{#090}{(x-2y)}}_{x^2-4y^2}\left.\!\vphantom{2^2}\right) $$

Answer 5

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 $a=1$.