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anonymous

  • 5 years ago

81s^6-3f^6 Factor

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  1. amistre64
    • 5 years ago
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    factor out the 3 and look for diff of squares

  2. amistre64
    • 5 years ago
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    or even diff of cubes

  3. amistre64
    • 5 years ago
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    27 is a cube so...

  4. amistre64
    • 5 years ago
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    3(27s^6 - f^6) 3 (3s^2 -f^2) (9s^4 +3(sf)^2 +f^4) if I remember correctly; and then theres that diff of squares that it generated....

  5. amistre64
    • 5 years ago
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    you have any with sqrt(3) in it?

  6. amistre64
    • 5 years ago
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    3 (sqrt(3)s +f)(sqrt(3)s -f) (9s^4 +3(sf)^2 +f^4) but let me check thru that again

  7. anonymous
    • 5 years ago
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    not they all have ^6

  8. amistre64
    • 5 years ago
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    all the answers?

  9. smurfy14
    • 5 years ago
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    3(3s^2-f^2)(9s^4+3f^2s^2+f^4)

  10. amistre64
    • 5 years ago
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    3(27s^6 -f^6) is what it initially factor to

  11. amistre64
    • 5 years ago
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    but thats a difference of cubes

  12. anonymous
    • 5 years ago
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    ok so what shall I put in for the answer then please

  13. smurfy14
    • 5 years ago
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    \[3(3s^2-f^2)(9s^4+3f^2s^2+f^4)\] is the answer

  14. amistre64
    • 5 years ago
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    if we factor this completely i get: 3 .... yeah, like that again

  15. amistre64
    • 5 years ago
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    but, that (3s^2 -f^2) factors again

  16. anonymous
    • 5 years ago
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    so do u amistre64 agree with smurfy answer

  17. amistre64
    • 5 years ago
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    \[3(\sqrt{3}+f)(\sqrt(3)-f)(9s^4 +3s^2 f^2 +f^2)\]

  18. anonymous
    • 5 years ago
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    3(3s2−f2)(9s4+3f2s2+f4)

  19. amistre64
    • 5 years ago
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    that (3) is a typo; just sqrt(3) like the other one lol

  20. anonymous
    • 5 years ago
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    This is the answer: 3(3√+f)((√3)−f)(9s4+3s2f2+f2)

  21. amistre64
    • 5 years ago
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    i just dont know what your material is looking for for an acceptable answer with that

  22. amistre64
    • 5 years ago
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    \(\sqrt{3}\) is a valid number as far as i can tell

  23. anonymous
    • 5 years ago
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    it just say factor sir

  24. anonymous
    • 5 years ago
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    I think i'll learn partial differential equations before she learns how to factor

  25. amistre64
    • 5 years ago
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    my gut says to go all the way ;) \(3(\sqrt{3}+s)(\sqrt{3}-s)(9s^4+3s^2 f^2+f^4)\)

  26. anonymous
    • 5 years ago
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    ok i will try it

  27. amistre64
    • 5 years ago
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    ack..cant type lol

  28. anonymous
    • 5 years ago
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    lordamercy. isn't this just the difference of two cubes?

  29. amistre64
    • 5 years ago
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    make it: \(3(\sqrt{3}s-f)(\sqrt{3}s+f)(9s^4 +3s^2 f^2 +f^4)\)

  30. amistre64
    • 5 years ago
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    thats my final offer lol

  31. amistre64
    • 5 years ago
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    its a diff of cubes with a diff of squares snucked in

  32. smurfy14
    • 5 years ago
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    im telling you mine is correct you can check it if you dont believe me

  33. anonymous
    • 5 years ago
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    3(3√+f)((√3)−f)(9s4+3s2f2+f2)

  34. amistre64
    • 5 years ago
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    checking a partial simplification doesnt prove nuthin lol

  35. anonymous
    • 5 years ago
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    amistre64 is smarter!

  36. amistre64
    • 5 years ago
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    at times.... other times im just an idiot in disguise ;)

  37. anonymous
    • 5 years ago
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    haha

  38. anonymous
    • 5 years ago
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    ok smurfy you were right sorry

  39. smurfy14
    • 5 years ago
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    thank you :)

  40. amistre64
    • 5 years ago
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    yay!! smurfy

  41. amistre64
    • 5 years ago
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    even tho i typed it in like that first ...... lol

  42. anonymous
    • 5 years ago
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    \[3(27s^6-f^6)=3((s^2)^3-(f^2)^3\]\] \[a^3-b^3=(a-b)(a^2+ab+b^2)\] so you get \[3((3s^2)-f^2)((3s^2)^2+3s^2f^2+(f^2)^2)\] \[3(3s^2-f^2)(9s^4+3s^2f^2+f^4)\]

  43. anonymous
    • 5 years ago
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    hello amistre!

  44. amistre64
    • 5 years ago
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    yes; and the question I had was do we take the diff of squares further?

  45. amistre64
    • 5 years ago
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    howdy! :)

  46. anonymous
    • 5 years ago
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    ahh i see . first term is difference of two squares, if you are factoring over reals. yes.

  47. amistre64
    • 5 years ago
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    thats what I thought .....

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