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Simple math problems

  1. May 13, 2004 #1
    Tonight my friend came to me with some simple math problems. Most of them were easy, but some of them I could not do, or could not remember how to do. It is very frustrating because I should know how to do this since I did it years ago.

    I was wondering if you guys could help. The two problems I cant get are:

    1) [tex] xy{\sqrt{125x^3y^5}} - {\sqrt{12x^5y^7}} + x^2y^3{\sqrt{{80xy}} [/tex]

    2) [tex] {\sqrt[3]{16x^4y^3}} - 5x{\sqrt[3]{54xy^3}}} + 20y{\sqrt[3]{2x^4}} [/tex]

    It seems so easy, jjust adding and subtracting, but I don't remember how to do this type of stuff. :frown:

    Thanks in advance guys.
  2. jcsd
  3. May 13, 2004 #2
    [tex] xy{\sqrt{125x^3y^5}} - {\sqrt{12x^5y^7}} + x^2y^3{\sqrt{{80xy}} [/tex]

    Let's do the first term.


    Any powers of 2 can be brought out, so let's write it in terms of powers of 2:


    Bring out the squares




    Rinse and repeat for the remaining terms.

    And for the cube root problem, powers of 3 can come out. So group them into powers of 3.

  4. May 13, 2004 #3
    Thank you Cookiemonster.

    I still can't get it though. :redface:
  5. May 13, 2004 #4


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    [tex] {\sqrt[3]{16x^4y^3}} - 5x{\sqrt[3]{54xy^3}}} + 20y{\sqrt[3]{2x^4}} [/tex]

    16= 8*2 and 8= 23
    x4= x*x3

    so the first term is [itex]2xy\sqrt[3]{2x}[/itex]

    54= 2*27= 2*33
    so the second term is [itex]5x(3y)\sqrt[3]{2x}[/itex]

    In the third term x4= x*x3
    so the third term is [itex]20y(x)\sqrt[3]{2x}[/itex]

    That is: [itex]2xy\sqrt[3]{2x}+5x(3y)\sqrt[3]{2x}+20y(x)\sqrt[3]{2x}[/itex]

    Notice the [itex]\sqrt[3]{2x}[/itex] in each term?

    This is the same as [itex](2xy+ 15xy+ 20xy)\sqrt[3]{2x}[/itex]
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