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Math Help - Fast help needed

  1. #1
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    Jul 2009
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    Fast help needed

    Hello... I am Adnan from Bosnia... I am preparing for exam to be accepted in college ...
    I need help for this problem...

    x+y+z=1; <br />
x^2+y^2+z^2=1; <br />
x^3+y^3+z^3=1;
    Solve..
    Thanks
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  2. #2
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    Mar 2009
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    Alberta
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    Quote Originally Posted by c3nar1us View Post
    Hello... I am Adnan from Bosnia... I am preparing for exam to be accepted in college ...
    I need help for this problem...

    x+y+z=1; <br />
x^2+y^2+z^2=1; <br />
x^3+y^3+z^3=1;
    Solve..
    Thanks
    Because we can see that an increase in degree does not change the value of the answer, we know that the variables MUST be something that always stays the same with changing whole number powers. Which could be \pm1 or 0. But because a) the answer is 1 (So it is obvious the variables cannot be zero), and b) the answer is ALWAYS 1, we know that the variables must be positive. (If they were negative they would switch between negative and positive between the square and cube functions, which they don't).

    Using the above information, we can see that 1 variable must be 1, while the other two variables must be 0.
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  3. #3
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    Quote Originally Posted by Kasper View Post
    Because we can see that an increase in degree does not change the value of the answer, we know that the variables MUST be something that always stays the same with changing whole number powers. Which could be \pm1 or 0. But because a) the answer is 1 (So it is obvious the variables cannot be zero), and b) the answer is ALWAYS 1, we know that the variables must be positive. (If they were negative they would switch between negative and positive between the square and cube functions, which they don't).

    Using the above information, we can see that 1 variable must be 1, while the other two variables must be 0.
    Y but i need mathematical proof... i know that 1 is only possible solvation..
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  4. #4
    Super Member
    Joined
    Jan 2009
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    715
    Suppose  x,y,z ~~ are the roots of a cubic equation

    we have x+y+z = 1 so the equation is <br />
u^3 - u^2 + bu + c = 0

    At the same time ,

     x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+zx) = 1 so
     1 - 2(xy+yz+zx) = 1 \implies xy+yz+zx = 0

    The equation :  u^3 - u^2 + c = 0
    And

     x^3 = x^3 - x^2 + c  + ( x^2 - x + \frac{c}{x}) - c + x - \frac{c}{x} = 0+0 - c + x - \frac{c}{x} = x - c - \frac{c}{x}

    So  x^3 + y^3 + z^3 = x+y+z - c-c-c - c (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 1 - 3c - c [ \frac{ xy+yz+zx}{xyz}] = 1-3c

    It is given that  x^3 + y^3 + z^3 = 1
    Therefore ,  1 = 1-3c \implies c = 0

    The equation is u^3 - u^2 = 0
    u = 0 , 0, 1

    Therefore , (x,y,z) = (0,0,1),(0,1,0),(1,0,0)

    If the values of the right hand sides are not 1,the above method can be applied to solve it , It is not just used to prove the set of the solution is (0,0,1) in this case .
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  5. #5
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    Quote Originally Posted by simplependulum View Post
    Suppose  x,y,z ~~ are the roots of a cubic equatiox^3n

    we have x+y+z = 1 so the equation is <br />
u^3 - u^2 + bu + c = 0

    At the same time ,

     x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+zx) = 1 so
     1 - 2(xy+yz+zx) = 1 \implies xy+yz+zx = 0

    The equation :  u^3 - u^2 + c = 0
    And

     x^3 = x^3 - x^2 + c  + ( x^2 - x + \frac{c}{x}) - c + x - \frac{c}{x} = 0+0 - c + x - \frac{c}{x} = x - c - \frac{c}{x}

    So  x^3 + y^3 + z^3 = x+y+z - c-c-c - c (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 1 - 3c - c [ \frac{ xy+yz+zx}{xyz}] = 1-3c

    It is given that  x^3 + y^3 + z^3 = 1
    Therefore ,  1 = 1-3c \implies c = 0

    The equation is u^3 - u^2 = 0
    u = 0 , 0, 1

    Therefore , (x,y,z) = (0,0,1),(0,1,0),(1,0,0)

    If the values of the right hand sides are not 1,the above method can be applied to solve it , It is not just used to prove the set of the solution is (0,0,1) in this case .
    Nice.. thanks : )
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