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mazehaq Group Title

3(x-3)^1/3 - 4(x-3)^4/3 + (x-3)^7/3 = 0 To make it easier, I substituted (x-3) with x, in which I got x^1/3(3 - 4x +x^2) ...but it can't be factored. Help me out! It's a fairly easy question I bet, but I need a nice way of solving it. Thank you! :)

  • one year ago
  • one year ago

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  1. lgbasallote Group Title
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    actually you might want to substitute x - 3 with a instead to avoid confusion

    • one year ago
  2. lgbasallote Group Title
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    but you're right anyway... except for the last part

    • one year ago
  3. lgbasallote Group Title
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    wait on second thought you're right

    • one year ago
  4. lgbasallote Group Title
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    and it can be factored... \[\huge a^{\frac 13} (3 - 4a + a^2) \implies a^{\frac 13} (a^2 - 4a + 3)\] can you see it now?

    • one year ago
  5. lgbasallote Group Title
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    a^2 - 4a + 3 is factorable

    • one year ago
  6. mazehaq Group Title
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    Oh okay! I kept thinking that it was a^2 + 4a - 3

    • one year ago
  7. lgbasallote Group Title
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    oh. no wonder

    • one year ago
  8. lgbasallote Group Title
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    yeah. that one isn't factorable

    • one year ago
  9. mazehaq Group Title
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    so 3 - 4a + a^2 is a^2 - 4a + 3 ...right?

    • one year ago
  10. lgbasallote Group Title
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    yes

    • one year ago
  11. mazehaq Group Title
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    Okie dokie thank you!

    • one year ago
  12. lgbasallote Group Title
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    welcome

    • one year ago
  13. mazehaq Group Title
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    So wait, how would I figure out the final answer, I am at (x-3)^1/3 (x-7)(x-7)=0

    • one year ago
  14. lgbasallote Group Title
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    they shouldn't be equal...

    • one year ago
  15. lgbasallote Group Title
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    how did you factor a^2 - 4a + 3?

    • one year ago
  16. mazehaq Group Title
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    I skipped a few steps. Remember a = (x-3) a^1/3 (a-1)(a-3) = 0 (x-3)^1/3 [(x-3-1)(x-3-3)] = 0 (x-3)^1/3 (x-4)(x-6) = 0

    • one year ago
  17. mazehaq Group Title
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    sorry about my previous comment tho, I totally typed it wrong, my computer is lagging so bad.

    • one year ago
  18. lgbasallote Group Title
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    right. that's better

    • one year ago
  19. mazehaq Group Title
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    So how do I solve the end?

    • one year ago
  20. mazehaq Group Title
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    (x-3)^1/3 (x-4)(x-6) = 0

    • one year ago
  21. lgbasallote Group Title
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    that s the end

    • one year ago
  22. lgbasallote Group Title
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    isn't it?

    • one year ago
  23. lgbasallote Group Title
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    or do you want x?

    • one year ago
  24. mazehaq Group Title
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    don't I make each = to zero

    • one year ago
  25. mazehaq Group Title
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    Like, 2 of my answers will be 4 & 6...

    • one year ago
  26. lgbasallote Group Title
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    right

    • one year ago
  27. lgbasallote Group Title
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    now..just equate (x-3)^1/3 to 0

    • one year ago
  28. mazehaq Group Title
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    I know but how :(

    • one year ago
  29. lgbasallote Group Title
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    \[(x - 3)^{\frac 13} = 0\]

    • one year ago
  30. lgbasallote Group Title
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    take the cube root of both sides..what happens?

    • one year ago
  31. mazehaq Group Title
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    x-3=0?

    • one year ago
  32. lgbasallote Group Title
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    yes

    • one year ago
  33. mazehaq Group Title
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    therefore x = 3?

    • one year ago
  34. lgbasallote Group Title
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    now solve for x

    • one year ago
  35. lgbasallote Group Title
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    right

    • one year ago
  36. lgbasallote Group Title
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    so there's your answer

    • one year ago
  37. mazehaq Group Title
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    Okay, so answer is 3,4,6 :) thanks!

    • one year ago
  38. lgbasallote Group Title
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    yup and welcome

    • one year ago
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