Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

sasogeek

\(\large (a^2-b^2) \) 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

  • one year ago
  • one year ago

  • This Question is Closed
  1. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  2. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  3. lalaly
    Best Response
    You've already chosen the best response.
    Medals 0

    (a+bi)(a-bi) ?

    • one year ago
  4. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

    where from the i?

    • one year ago
  5. lalaly
    Best Response
    You've already chosen the best response.
    Medals 0

    do u know complex numbers?

    • one year ago
  6. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  7. lalaly
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  8. lalaly
    Best Response
    You've already chosen the best response.
    Medals 0

    because u said terms without exponents

    • one year ago
  9. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

    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...

    • one year ago
  10. lalaly
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah yeah

    • one year ago
  11. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  12. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

    For the cube thingy I don't know xD

    • one year ago
  13. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

    okay thanks for this one :)

    • one year ago
  14. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  15. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  16. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

    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 :|

    • one year ago
  17. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  18. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • one year ago
  19. sasogeek
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  20. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

    I think I got it :D

    • one year ago
  21. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

    \(\large a^3+b^3\) 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 \(\large (a+b)(a^2-ab+b^2)\)

    • one year ago
  22. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • one year ago
  23. zepp
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  24. across
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.