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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 :)

Nicolas Leskiu
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  • You might try to apply the formula for the sum of 2 cubes, $x^3+y^3= (x+y)(x^2-xy+y^2)$. – sharding4 May 23 '17 at 19:19
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    Hint: Have you tried to simply expand $(a^4-2ab^3)^3+(2a^3b-b^4)^3$? You should get something in terms of $x=a^3$ and $y=b^3$ which you can factor as $(x-y)(x+y)^3$. – fractal1729 May 24 '17 at 02:47
  • @fractal1729 I did that, and found nothing, am I losing something? – Nicolas Leskiu May 24 '17 at 02:49
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    Do. Not. Reask. Questions. EVER!!! That site rule is strictly enforced. The interested answerers have zero trouble finding the first edition, so there is absolutely NO POINT in asking the same question twice. You only create orphaned garbage questions by doing that. Furthermore, some questions may take days or weeks to get an answer (if ever). It is NOT as if that when they fall out of the front page they would become inaccessible. – Jyrki Lahtonen May 24 '17 at 05:12
  • @JyrkiLahtonen Sorry about that, thanks for the advice. Is the other question deleted? I can't see if it is. – Nicolas Leskiu May 25 '17 at 01:55
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    Yup. I merged the other question with this so the answers came here as well (we the diamond moderators have access to such tools). I then deleted it, because there was no point in keeping that stub around. No worries. New users don't always no about this rule. I guess it is explained somewhere in the help center, but it is not advertised well enough. – Jyrki Lahtonen May 25 '17 at 06:01

3 Answers3

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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?

mweiss
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HINT:

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

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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 $a^3-b^3$.

csar
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