## sasogeek 3 years ago $$\large (a^2-b^2)$$ is proven to be equal to $$\large (a+b)(a-b)$$ (difference of two squares) . what then is $$\large (a^2+b^2)$$ and hence expand $$\large (a^3+b^3)$$ to the same form without the exponents. P.S. this is my own personal question. I need some understanding

1. sasogeek

by expansion to "the same form without the exponents", i mean, expand it to a form that does not have terms with exponents

2. zepp

I guess for the $$\large a^2 + b^2$$ you can use the complete the square method, let me explain down there.

3. lalaly

(a+bi)(a-bi) ?

4. sasogeek

where from the i?

5. lalaly

do u know complex numbers?

6. sasogeek

i've heard of them... not sure if i invited them to this party? how did they get here?

7. lalaly

lol thats how i understood ur problem, lol never mind then

8. lalaly

because u said terms without exponents

9. sasogeek

i said that because difference of two squares reduces to $$(a+b)(a-b)$$ and there's no exponents in there... if u see what i mean...

10. lalaly

yeah yeah

11. zepp

\large \begin{align} a^2+b^2 \\ a^2+b^2 +2ab-2ab \\ (a+b)^2-2ab \\ \text{Difference of squares} \\ (a+b+\sqrt{2ab})(a+b-\sqrt{2ab})\end{align} There :)

12. zepp

For the cube thingy I don't know xD

13. sasogeek

okay thanks for this one :)

14. zepp

Wait, I think I have a pdf file somewhere, talking about factorisation, wait a second let me find it.

15. sasogeek

does this work? $$\ a^3+b^3$$=(a+b)(a+b)(a+b)

16. zepp

I don't think so, I couldn't find it but I think you'll have to expand this into a 2nd degree form with some trashes behind :|

17. sasogeek

lol ok i'll see how best i can handle this. i'm sure there's something to it that works :) thanks anyway :)

18. zepp

I browsed through the web and found that would be (a+b)(a^2-2ab+b^2), I don't know the steps though :( and I guess your first step was right, but with a minus sign somewhere lol

19. sasogeek

but a^2-2ab+b^2 is the same as (a-b)^2 :/

20. zepp

I think I got it :D

21. zepp

$$\large a^3+b^3$$ Since it's a cube, I'll need a^2b and b^2a, I can't really explain why I chose them but :P (it was instinct lol) $$\large a^3+a^2b-a^b+b^2a-b^2a +b^3$$ From this, I can factor out all similar stuffs It would become (I'll rearrange that first) $$\large a^3-a^2b+ab^2+a^2b-ab^2+b^3$$ There Now $$\large a(a^2-ab+b^2)+b(a^2-ab+b^2)$$ Thus $$\large (a+b)(a^2-ab+b^2)$$

22. zepp

I guess I chose $$\large a^2b ~~\text{and}~~ b^2a$$ because I'll have to reduce a^3 and b^3 to squares, a^2 and b^2 Then since there's 3 a, 3 b, I add b and a, a^2b and b^2a

23. zepp

http://puu.sh/uaUs Should be a^2b

24. across

zepp is right on. It's easy to think that$(a^n+b^n)=(a+b)^n\\\exists a,b,n\in\mathbb{Z}.$This is not true.