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Easy Question!

  1. Dec 5, 2007 #1
    In the middle of doing an integral for Calc 3, and I came across (9-3x)^3, and completely forgot how to do this from algebra! Help!
     
  2. jcsd
  3. Dec 5, 2007 #2

    rock.freak667

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    Use u=9-3x
    du=-3 dx so that -du/3 = dx

    so you must really integrate

    [tex]\int \frac{-1}{3}u^3 du[/tex]
     
  4. Dec 5, 2007 #3
    Oh, I guess I didn't phrase my question right, huh? Haha. I'm doing an integral, yes, but I'm not actually taking the integral of that... I just need to simply that part of a loooong answer I got inside an integral..... if that makes sense. Just algebra stuff.
     
  5. Dec 5, 2007 #4
    Is this what it is:

    (a + b)^3 = a3 + 3a^2b + 3ab^2 + b^3

    ???
     
  6. Dec 6, 2007 #5

    rock.freak667

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    yes that is the expansion of (a+b)^3
     
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