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lokiando

  • one year ago

Which product will result in a sum or difference of cubes? a) (x + 7)(x^2 – 7x + 14) b) (x + 8)(x^2 + 8x + 64) c) (x – 9)(x^2 + 9x + 81) d) (x – 10)(x^2 – 10x + 100)

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  1. lokiando
    • one year ago
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    @UsukiDoll

  2. UsukiDoll
    • one year ago
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    sum of cubes formula \[(a+b)(a^2-ab+b^2)\] difference of cubes formula \[(a-b)(a^2+ab+b^2)\]

  3. lokiando
    • one year ago
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    c?

  4. lokiando
    • one year ago
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    will give medal

  5. anonymous
    • one year ago
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    I think the answer is C.

  6. lokiando
    • one year ago
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    you think? mmmmm

  7. anonymous
    • one year ago
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    Not really sure on this one. I was thinking either A. or C.

  8. UsukiDoll
    • one year ago
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    ah I see we have to go backwards on this.. so we need all perfect cubes. . no it can't be a.

  9. anonymous
    • one year ago
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    Because it can't be factored right? @UsukiDoll

  10. UsukiDoll
    • one year ago
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    8 27 64 \[2^3 = 8 , 3^3 =27, 4^3 = 64, 5^3 = 125\]

  11. UsukiDoll
    • one year ago
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    @200205650 true and also 14 is not a perfect cube

  12. lokiando
    • one year ago
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    okay

  13. lokiando
    • one year ago
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    so its c?

  14. UsukiDoll
    • one year ago
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    sum of cubes \[(a^3+b^3) = (a+b)(a^2-ab+b^2) \] difference of cubes \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \]

  15. UsukiDoll
    • one year ago
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    these choices are wonky though

  16. anonymous
    • one year ago
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    I don't see how you can factor any of them in the answer choices...unless you use the quadratic formula.

  17. UsukiDoll
    • one year ago
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    like .. let a = x let b = 4 \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \] \[(x^3-4^3) = (x-4)(x^2+4x+16) \]

  18. UsukiDoll
    • one year ago
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    I see perfect squares for the last 3 choices that's true.. for the ending part is b^2 . It's like this question isn't following the rules. That's what's getting me mad

  19. lokiando
    • one year ago
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    same here

  20. UsukiDoll
    • one year ago
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    let a = x and let b = 9 difference of cubes formula \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \] \[(x^3-9^3) = (x-9)(x^2+9x+81) \]

  21. anonymous
    • one year ago
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    From your examples so far @UsukiDoll I would think C. is the right answer.

  22. UsukiDoll
    • one year ago
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    that is messed up... really. 9 isn't a perfect cube to begin with. perfect square yes but not perfect cube

  23. anonymous
    • one year ago
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    Ya it isn't...but the way the equations are set up B. and D. don't follow that same pattern.

  24. UsukiDoll
    • one year ago
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    only perfect cubes are allowed for these formulas...>:/ oh I'm missing something XD 9 x 9 x 9 = 729

  25. UsukiDoll
    • one year ago
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    729 is a perfect cube number. that guy is 9

  26. UsukiDoll
    • one year ago
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    ok I got it.. I missed a very important step let a = x and b = 9 \[(x^3-729) = (x-9)(x^2+9x+81)\] \[(x^3-9^3) = (x-9)(x^2+9x+81)\]

  27. anonymous
    • one year ago
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    So the answer is C.?

  28. UsukiDoll
    • one year ago
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    that 729 is rewritten as 9^3 in the second line so a = x (the variable) and b = 9 (the number) a^2 = x^2 ab = 9x b^2 = (9)(9) = 81

  29. UsukiDoll
    • one year ago
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    yeah C matches ... on top of that .. the signs match for the difference of cubes. I wasn't working backwards until the very beginning.

  30. anonymous
    • one year ago
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    Great job explaining that!

  31. lokiando
    • one year ago
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    well thank you guys for your time and patients :)

  32. UsukiDoll
    • one year ago
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    XD patience XDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

  33. UsukiDoll
    • one year ago
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    It's past my lunch time too... so that's why my thinking started to go off

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