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Some sums

  1. Nov 19, 2009 #1
    Hi there!!
    Here are a few sums that are making me go nuts:cry:( actually can't get any clue how to solve:confused:)
    so here they are
    ( a+x)^1/3 + (a-x)^1/3= b
    ( Gosh I wish there could be some rule so that we could straight away write a^ 1/3 and x^1/3 anyway:tongue:)
    And 1/p+q+x=1/p+1/q+1/x
    I WOULD BE RALLY VERY THANKFUL IF ANYONE COULD GIVE A "HINT:cry:"!!
    THANK YOU:smile:
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2

    Mark44

    Staff: Mentor

    When you have fractional exponents written as you have them, use parentheses around the exponent. The 1/3 power is the same as the cube root.

    [tex]\sqrt[3]{a + x} + \sqrt[3]{a + x}~=~2\sqrt[3]{a + x}[/tex]
    Now, divide both sides of your equation by 2, and then cube both sides.

    For you other problem, 1/p+q+x=1/p+1/q+1/x, it's hard to say anything without knowing exactly what the problem is -- use parentheses.
    As you have written it, the problem is (1/p) + q + x = 1/p+1/q+1/x. I suspect that you meant the left side to be 1/(p + q + x), but I'm not sure.
     
    Last edited: Nov 19, 2009
  4. Nov 19, 2009 #3
    Ya it's 1/(p+q+x)
     
  5. Nov 19, 2009 #4
    Sorry i got the sign wrong it should be
    ( a+x)^1/3 + (a-x)^1/3= b
    I am REALLY VERY SORRRRRY!!!
     
  6. Nov 19, 2009 #5

    Mark44

    Staff: Mentor

    That makes it a different, and harder, problem...
     
  7. Nov 19, 2009 #6
    Substitute x =a + t^3 to write this as:

    (2a + t^3)^1/3 - t = b ------->

    (2a + t^3)^1/3 = b + t

    Take the cube of both sides. You'll see that the t^3 term cancels, so you have a quadratic equation for t.
     
  8. Nov 22, 2009 #7
    Thanks A LOT . :smile::smile:
    I have figured the answer !!:biggrin:
     
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