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Cutiepo0

  • 2 years ago

How to factor (x^3 + 1)

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  1. Brittni0605
    • 2 years ago
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    \[(x^{3}+1)\]

  2. Cutiepo0
    • 2 years ago
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    mmhm!

  3. bandicoot12
    • 2 years ago
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    no, (x+1)(x^2-x+1)

  4. Hero
    • 2 years ago
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    \[(a + b)^3 = (a + b)(a^2 - ab + b^2)\]

  5. Brittni0605
    • 2 years ago
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    \[(x+1)(x^{2}-x+1)\]

  6. Hero
    • 2 years ago
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    That's the sum of cubes formula I posted above.

  7. bandicoot12
    • 2 years ago
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    ya basically what they said

  8. Cutiepo0
    • 2 years ago
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    ok but I just established that (x^3+1) and (x+1)^3 weren't the same thing :S

  9. Hero
    • 2 years ago
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    No, they are not. One is simply x cubed plus one. The other is \((x + 1) \times (x + 1) \times (x + 1)\)

  10. Cutiepo0
    • 2 years ago
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    right, so then how can I use the sum of cubes formula? b/c isn't that for (a+b)3

  11. Hero
    • 2 years ago
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    Because \((x^3 + 1) = (x^3 + 1^3)\)

  12. Hero
    • 2 years ago
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    Remember, \(1 = 1 \times 1 \times 1 = 1^3 \)

  13. Cutiepo0
    • 2 years ago
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    = (x + 1) ^3 ?? but wouldnt that make it the same??

  14. Hero
    • 2 years ago
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    No, I just explained to you earlier that \((x^3 + 1) \ne (x + 1)^3\)

  15. Hero
    • 2 years ago
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    \((x^3 + 1) = (x^3 + 1^3)\) \((x + 1)^3 = (x + 1)(x + 1)(x + 1)\)

  16. Cutiepo0
    • 2 years ago
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    ok, so (a^3 + b^3) doesn't equal (a+b)^3, ever right? Sorry, just trying to wrap my head around it

  17. Hero
    • 2 years ago
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    The only way is if x = 0 or x = 1

  18. Hero
    • 2 years ago
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    But in general, they will not be equal

  19. Cutiepo0
    • 2 years ago
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    ok, so now though with the sum of cubes formula, the whole point was to find the zeroes of the equation f(x)= (x^3 + 1)(x-17)

  20. Cutiepo0
    • 2 years ago
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    and the answer is x=17 and x = -1

  21. Cutiepo0
    • 2 years ago
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    so how do I get from the sum of cubes to -1

  22. Hero
    • 2 years ago
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    Basically, the question is asking: What x values would make \((x^3 + 1)(x - 17) = 0\) You know that anything times zero equals zero. So when you expand \((x^3 + 1)\) you get \((x + 1)(x^2 - x + 1)\) Thus \((x^3 + 1)(x - 17) = (x + 1)(x^2 - x + 1)(x - 17) = 0\)

  23. Hero
    • 2 years ago
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    If you look closely you'll realize that if (x + 1) = 0 or if (x - 17) = 0, then the product of the whole thing will be zero. There's only one way to make (x + 1) = 0 There's only one way to make (x - 17) = 0.

  24. Hero
    • 2 years ago
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    What I say doesn't matter if you don't understand anything.

  25. Hero
    • 2 years ago
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    The silence is deafening

  26. Cutiepo0
    • 2 years ago
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    Oh, I got it, thanks a lot for your help with this :) b/c (x2−x+1) doesn't have any roots, so it's just the other two roots

  27. Hero
    • 2 years ago
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    Oh goodie :D

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