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Algebra Problem, solving by rearranging

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Let x and y be real numbers with x+y=1 and (x2 + y2)(x3 + y3) = 12. What is the value of x2 + y2 ?


    2. Relevant equations
    Sum of Cubes: (a3 + b3) = (a+b)(a2-ab+b2)


    3. The attempt at a solution
    I plugged in the sum of cubes to the equation that equals 12 to get:
    (x2 + y2)(x+y)(x2-xy+y2) = 12
    and since (x+y) = 1,
    (x2 + y2)(x2-xy+y2) = 12
    and therefore,
    (x2 + y2)(x2-xy+y2) = (x2 + y2)(x3 + y3)
    cancellation:
    (x+y)(x2-xy+y2) = (x3 + y3) = 12

    then I plugged (x3 + y3) for 12 to the original equation:
    (x2 + y2)(x3 + y3) = (x3 + y3)
    and that means that (x2 + y2) = 1, right?
     
  2. jcsd
  3. Apr 3, 2012 #2
    The answer key that I just found said that the answer is 3...
    Where did I go wrong?
     
  4. Apr 3, 2012 #3

    SammyS

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    How did you arrive at the above line ?
     
  5. Apr 3, 2012 #4
    A typo, my bad,
    what I meant was:

    and therefore,
    (x2 + y2)(x+y)(x2-xy+y2) = (x2 + y2)(x3 + y3)
    cancellations:
    (x2 + y2)(x2-xy+y2) = (x2 + y2)(x3 + y3) = 12
    (x2-xy+y2) = (x3 + y3)

    I plugged in the sum of cubes to the equation that equals 12 to get:
    (x2 + y2)(x+y)(x2-xy+y2) = 12
    and since (x+y) = 1,
    (x2 + y2)(x2-xy+y2) = 12
    and therefore,
    (x2 + y2)(x2-xy+y2) = (x2 + y2)(x3 + y3)
    cancellation:
    (x2-xy+y2) = (x3 + y3)

    Now I realize that the last part is wrong, but where do I go from here?
     
  6. Apr 4, 2012 #5
    How did you get from

    (x2 + y2)(x2-xy+y2) = 12
    to
    (x2 + y2)(x+y)(x2-xy+y2) = (x2 + y2)(x3 + y3) ?

    I'm not saying it's wrong, I just don't follow it.

    Anyway, after several pages of algebra, I also arrived at x2+y2=1
    Are you sure you're looking at the right answer?

    [itex]x=1±\frac{1}{\sqrt{2}} , 1±\frac{\sqrt{5}i}{\sqrt{3}}[/itex]

    edit: I plugged it in to the original equation and it doesn't work, so I must have made a mistake somewhere. Odd that we both got the same answer..
     
    Last edited: Apr 4, 2012
  7. Apr 4, 2012 #6
    Let
    [tex] A = x^2 + y^2[/tex]

    [tex](x^2 + y^2)(x^3 + y^3) = 12[/tex]
    [tex]A(x+y) (x^2-x y+y^2)=12[/tex]
    [tex]A(A-x y)=12[/tex]
    [tex]x+y = 1 \rightarrow (x+y)^2 = 1 \rightarrow x^2+y^2+2xy = 1 \rightarrow xy = \frac{1-x^2-y^2}{2}=\frac{1-A}{2}[/tex]
    [tex]A(A-\frac{1-A}{2})=12[/tex]
    [tex]A(\frac{3}{2}A-\frac{1}{2})=12[/tex]
    [tex]3A^2-A-24=0[/tex]
    You get two answers, but one is negative. We know the answer must be positive. The answer is 3.
     
  8. Apr 4, 2012 #7

    verty

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    This is an olympiad type question, you need to play with it. Start with x = 1-y, then get formulas for (x^2 + y^2) and (x^3 + y^3). Look to make a substitution.
     
  9. Apr 4, 2012 #8
    thanks for your help! that makes perfect sense
     
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