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Solving equation with cube roots.

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the equation:

    [itex]\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0[/itex]


    3. The attempt at a solution

    What I did was move (x+3)^(1/3) to the other side, cube both sides and when I put them equal to 0 again, I managed to factor (x+2)^(1/3) out of it giving one solution x=-2.

    However if I looked up the proposed solution after cubing and a little gathering and grouping it arrived at this:

    [itex]3x+6=-3\sqrt[3]{x+1}*\sqrt[3]{x+2}*(\sqrt[3]{x+1}+\sqrt[3]{x+2})[/itex]

    Which is clear, but then in the next step it is converted to this without explanation:

    [itex]x+2=\sqrt[3]{(x+1)(x+2)(x+3)}[/itex]

    From there on the solution is just putting it equal to 0 and factorizing, but how did it get to this from the previous? It would imply [itex]-(\sqrt[3]{x+1}+\sqrt[3]{x+2})[/itex] is equal to [itex]\sqrt[3]{(x+3)}[/itex], which I don't think it is. Is it wrong or what is up with that?
     
  2. jcsd
  3. Jun 24, 2012 #2

    phyzguy

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    Science Advisor

    Look at your original equation:
    [tex]\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0[/tex]
    This means:
    [tex]\sqrt[3]{x+1}+\sqrt[3]{x+2}=-\sqrt[3]{x+3}[/tex]
    Plug this into:
    [tex]3x+6=-3\sqrt[3]{x+1}*\sqrt[3]{x+2}*(\sqrt[3]{x+1}+\sqrt[3]{x+2})[/tex]
    and cancel out a factor of 3 to give:
    [tex]x+2=\sqrt[3]{(x+1)(x+2)(x+3)}[/tex]
     
  4. Jun 24, 2012 #3
    Oh, didn't see that, thank you.
     
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