Factorization, simple expression!

Question

I must factor out this expression: (I don't speak English, I don't know the right mathematical terms you use in English, sorry if I make a mistake).

$a^{3}(\frac{a^{3}-2b^{3}}{a^{3}+b^{3}})^{3}+b^{3}(\frac{2a^{3}-b^{3}}{a^{3}+b^{3}})^{3}$

The answer is: $a^{3}-b^{3}$

This is what I found:

$\frac{[a(a^{3}-2b^{3})]^{3}}{(a^{3}+b^{3})^{3}}+\frac{[b(2a^{3}-b^{3})]^{3}}{(a^{3}+b^{3})^{3}}$

$\frac{1}{(a^{3}+b^{3})^{3}}([a(a^{3}-2b^{3})]^{3}+[b(2a^{3}-b^{3})]^{3})$

Now I don't know what to do, principally with this two terms: $(a^{3}-2b^{3})$ and $(2a^{3}-b^{3})$

I hope you help me, thanks :)

Answer 1

Just power through it: $$a^{3}\left(\frac{a^{3}-2b^{3}}{a^{3}+b^{3}}\right)^{3}+b^{3}\left(\frac{2a^{3}-b^{3}}{a^{3}+b^{3}}\right)^{3}$$

$$\frac{a^3(a^3-2b^3)^3+b^3(2a^3-b^3)^3}{(a^3+b^3)^3}$$

$$\frac{a^3(a^9-6a^6b^3+12a^3b^6-8b^9)+b^3(8a^9-12a^6b^3+6a^3b^6-b^9)}{(a^3+b^3)^3} $$

$$\frac{a^{12}-6a^9b^3+12a^6b^6-8a^3b^9+8a^9b^3-12a^6b^6+6a^3b^9-b^{12}}{(a^3+b^3)^3} $$

$$\frac{a^{12}+2a^9b^3-2a^3b^9-b^{12}}{(a^3+b^3)^3} $$

$$\frac{(a^6+2a^3b^3+b^6)(a^6-b^6)}{(a^3+b^3)^3} $$

Can you finish it from there?

Answer 2

HINT:

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

Answer 3

Let $x=a(a^3-2b^3)$ and $y=b(2a^3-b^3)$.

Use the identity $x^3+y^3=(x+y)(x^2-xy+y^2)$ and you will get (after some manipulations):

• $x+y = (a-b)(a+b)^3$
• $x^2-xy+y^2 = (a^2-ab+b^2)^3(a^2+ab+b^2)$.

Now put everything together and you have $a^3-b^3$.