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Viete relations

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider the polynomials:

    f(x) = x[itex]^{6}[/itex] + x[itex]^{3}[/itex] +1 and g(x) = x[itex]^{2}[/itex] + x + 1

    Denote the roots of f(x) = 0 by x[itex]_{1}[/itex], ... , x[itex]_{6}[/itex].

    Show that [itex]\sum[/itex]g(x[itex]_{k}[/itex]) = 6 , 1[itex]\leq[/itex]k[itex]\leq[/itex]6


    2. Relevant equations

    Vieta relations.

    3. The attempt at a solution

    Please correct any initial mistakes I may have made in this, new to Vieta.

    [itex]\sum[/itex]x[itex]_{i}[/itex] = 0, 1[itex]\leq[/itex]i[itex]\leq[/itex]6

    [itex]\sum[/itex]x[itex]_{i}[/itex]x[itex]_{j}[/itex]x[itex]_{k}[/itex] = -1, 1[itex]\leq[/itex]i,j,k[itex]\leq[/itex]6, i[itex]\neq[/itex]j[itex]\neq[/itex]k

    and

    [itex]\prod[/itex]x[itex]_{i}[/itex] = 1, 1[itex]\leq[/itex]i[itex]\leq[/itex]6.

    Then we must show that

    x[itex]^{2}_{1}[/itex] + x[itex]_{1}[/itex] + 1 + x[itex]^{2}_{2}[/itex] + x[itex]_{2}[/itex] + 1 + x[itex]^{2}_{3}[/itex] + x[itex]_{3}[/itex] + 1 + x[itex]^{2}_{4}[/itex] + x[itex]_{4}[/itex] + 1 + x[itex]^{2}_{5}[/itex] + x[itex]_{5}[/itex] + 1 + x[itex]^{2}_{6}[/itex] + x[itex]_{6}[/itex] + 1 = 6

    or using the first vieta relation we must show that

    x[itex]^{2}_{1}[/itex] + x[itex]^{2}_{2}[/itex] + x[itex]^{2}_{3}[/itex] + x[itex]^{2}_{4}[/itex] + x[itex]^{2}_{5}[/itex] + x[itex]^{2}_{6}[/itex] = 0

    Then I'm stuck, if I did everything right so far.

    Never mind, I think I might have gotten it, I could just multiply the first vieta relation by itself to get what I need, the x's square, which will include the second two vieta relations which'll add up to zero and all the other combinations of the x's will automatically be zero since we had no coefficients for some of the polynomial degrees, right?
     
    Last edited: Dec 2, 2011
  2. jcsd
  3. Dec 2, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Right, I think. You didn't really show the whole thing. But sure, if you square the first relation you'll get the sum of the squares plus something else, which you can also show to be zero.
     
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