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Math Help - Can anyone help with this problem?

  1. #1
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    Can anyone help with this problem?

    cant seem to figure out how to do this problem

    Suppose x+y= 1 and x^2+y^2=4

    find with algebraic explaination the value of x^3+y^3
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mcdaking84 View Post
    cant seem to figure out how to do this problem

    Suppose x+y= 1 and x^2+y^2=4

    find with algebraic explaination the value of x^3+y^3
    we can use the sum of two cubes formula:

    Note that x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)(x^2 + y^2 - xy)

    now we can plug in x+y= 1 and x^2+y^2=4

    so x^3 + y^3 = (1)(4 - xy)
    => x^3 + y^3 = 4 - xy

    if we wanted to, we could solve for x and y directly, but the question didn't ask for that

    EDIT: Oh sorry, we're not done yet. We need to find the value of x^3 + y^3, so we don't want xy in the expression. let's find a numerical value for xy.

    we are told x + y = 1
    => (x + y)^2 = 1^2
    => x^2 + 2xy + y^2 = 1 .........but x^2 + y^2 = 4
    => 4 + 2xy = 1
    => 2xy = -3
    => xy = -3/2

    so x^3 + y^3 = 4 - xy = 4 + 3/2 = 11/2 or 5 1/2
    Last edited by Jhevon; April 19th 2007 at 05:37 PM.
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  3. #3
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    Hello, mcdaking84!

    Another approach . . .


    Given: . .x + y .= .1 - [1]
    . .and: .x + y .= .4 .[2]

    Find the value of: .x + y

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

    From [2], we have: .2xy + 4 .= .1 . . xy = -3/2 .[3]


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

    Then we have: . . . x + y + 3xy(x + y) .= .1
    . . . . . . . . . . . . . . . . . . . . . . . . .
    From [3] and [1]: .x + y + 3(-3/2)(1) .= .1


    Therefore: .x + y - 9/2 .= .1 . . x + y .= .11/2

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