Results 1 to 3 of 3

Math Help - Square expresion

  1. #1
    Junior Member
    Oct 2005

    Square expresion

    It seems that the part of my last post that stated my tight sleep as true was indeed a lie. I now have another problem witch will haunt my dreems into infinity, that is iff someone doesn't help me a little!

    The problem is:
    Simplify this as far as possible.

    As for my attempt of a solution:





    Thus the original expression equals:
    3*\sqrt[3]{3}-\sqrt[3]{3}+2\sqrt{3}-\sqrt{3}-3^{4/9} = }
    2*\sqrt[3]{3}+\sqrt{3}-3^{4/9} \not= 2*\sqrt[3]{3}

    Where 2*\sqrt[3]{3} is the answer given by my textbook.
    What did I do wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    May 2006
    New England
    Quote Originally Posted by a4swe
    This is where you're off, it should be:


    Now let's finish: \sqrt[3]{81}-\sqrt[6]{9}+\sqrt{12}-\sqrt[6]{27}-\sqrt[3]{3\sqrt3}

    simplify: 3*\sqrt[3]{3}-\sqrt[3]{3}+2\sqrt{3}-\sqrt{3}-\sqrt3

    group like terms: \left(3*\sqrt[3]{3}-\sqrt[3]{3}\right)+\left(2\sqrt{3}-\sqrt{3}-\sqrt3\right)

    simplify: \left((3-1)*\sqrt[3]{3}\right)+\left(0\right)

    therefore: \boxed{2*\sqrt[3]{3}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Oct 2005
    Thank you very much, that was quick.
    I thought it was some minor error just like that witch I just couldn't spot.
    And belive me, I have looked for it, for a long time.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 2nd 2011, 08:03 PM
  2. perfect square trinomial/square of binomial
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 3rd 2011, 05:02 PM
  3. Replies: 1
    Last Post: November 24th 2009, 09:17 AM
  4. Replies: 11
    Last Post: October 25th 2009, 07:45 PM
  5. [SOLVED] Deriving a square root under a square
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 29th 2007, 07:21 AM

Search Tags

/mathhelpforum @mathhelpforum