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Math Help - x^3 - y^3

  1. #1
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    x^3 - y^3

    Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.

    I remember I knew how to solve this at one point, but forgot over the summer.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ceasar_19134 View Post
    Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.

    I remember I knew how to solve this at one point, but forgot over the summer.
    x-y=2, now square both sides to get:

    (x-y)^2=x^2+y^2-2xy=8-2xy=2^2

    so xy=2.

    Now multiply x-y and x^2+y^2 together to get:

    (x-y)(x^2+y^2)=x^3-y^3+xy^2-yx^2=(x^3-y^3) + xy(y-x)

    But the the term at the extreme left is equal to 2.8=16, and the term at the
    extreme right is (x^3-y^3)-4, hence x^3-y^3=20.

    RonL

    Please check the algebra carefully to make sure it's correct.
    Last edited by CaptainBlack; October 2nd 2006 at 07:49 AM. Reason: to correct algebra
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  3. #3
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    Quote Originally Posted by ceasar_19134 View Post
    Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.

    I remember I knew how to solve this at one point, but forgot over the summer.
    Hi,

    compare (x-y)^2 = x^2 - 2xy + y^2 = 4 and x^2 + y^2 = 8, thus 2xy = 4, thus xy = 2.

    From x-y = 2 you get: x = y+2, therefore xy = 2 becomes: (y+2)y = 2

    Simplify to: y^2 + 2y - 2 = 0

    I leave the next steps to you. Substitute y in the equation x - y = 2 to calculate x.

    I've got as final result to your problem x^3 - y^3 = 20.

    tschüss

    EB
    Last edited by earboth; October 2nd 2006 at 06:54 AM. Reason: I found an ugly mistake
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  4. #4
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    Hello, ceasar_19134!

    A variation of the Captain's solution . . . and Earboth's.


    Suppose that x and y are two real numbers such that:
    . . [1] x - y .= .2 .and .[2] x + y .= .8

    Find: x - y

    Square [1]: .(x - y) .= .2 . . x - 2xy + y .= .4 . . (x + y) - 2xy .= .4

    . . From [2] we have: .8 - 2xy .= .4 . . -2xy = -4 . . xy = 2 .[3]


    Cube [1]: . (x - y) .= .2 . . x - 3xy + 3xy - y .= .8

    . . We have: . x - y - 3xy(x - y) .= .8


    From [2] and [3], we have: . x - y - 3(2)(2) .= .8


    Therefore: . x - y .= .20

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  5. #5
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    whats the next step after y^2+2y-2=o
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  6. #6
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    Quote Originally Posted by Rimas View Post
    whats the next step after y^2+2y-2=o
    Hi, that's a quadratic equation. Use the formula, which you probably know, to solve this equation:

    If the equation is: ax^2 + bx + c = 0 then the solution is: x = 1/(2a)(-b+- sqrt(b^2-4ac))

    Use this formula: y= 1/(2*1)(-2+- sqrt(2^2-4*1*(-2)) = -1+- sqrt(3)

    tschss

    EB
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