text: ## 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.
explanation: The document contains a math question and a series of responses attempting to answer the question. The user asks about the expansion of \(a^2 + b^2\) and \(a^3 + b^3\) into forms without exponents. The responses provide different approaches and explanations, including the use of complex numbers and completing the square for \(a^2 + b^2\), and a detailed step-by-step expansion for \(a^3 + b^3\).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a series of responses attempting to answer the question. The user asks about the expansion of \(a^2 + b^2\) and \(a^3 + b^3\) into forms without exponents. The responses provide different approaches and explanations, including the use of complex numbers and completing the square for \(a^2 + b^2\), and a detailed step-by-step expansion for \(a^3 + b^3\).
  
  Label: question-and-answer