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    jim556's Avatar
    jim556 Posts: 8, Reputation: 1
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    #1

    Apr 27, 2008, 12:15 AM
    having trouble solving precalculus equation
    hi, I have been having some trouble doing this problem, could some one help me out?
    If a+b=6, ab=4 find the value of a^3+b^3. (a+b)^3=a^3+3a^2b+3ab^2+b^3
    I'm sorry if this is a bit unclear, the a^2 means a to the power of 2. thanks for the help.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Apr 27, 2008, 05:13 AM
    Perhaps, try using the sum of two cubes factorization.



    Then we get it down to

    Did they give you by chance?
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    jim556 Posts: 8, Reputation: 1
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    #3

    Apr 27, 2008, 09:08 AM
    No they didn't, sorry, that was all that it said
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #4

    Apr 27, 2008, 09:47 AM
    Then we can finish by solving for a or b in the givens.

    ab=4... [1]

    a+b=6... [2]

    From [1], a=4/b

    Sub into [2]:

    (4/b)+b=6



    Now, can you finish up?
    jim556's Avatar
    jim556 Posts: 8, Reputation: 1
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    #5

    Apr 27, 2008, 07:50 PM
    Yes, thank you for the help Galactus.
    alamleh's Avatar
    alamleh Posts: 8, Reputation: 3
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    #6

    Apr 28, 2008, 08:29 AM
    you may finish the solution #2 suggested by galacuts like this:
    (a+b)^2=a^2 + b^2 + 2ab
    36 = a^2+b^2 + 8 , therefore, a^2+b^2=28

    now a^3+b^3 =(a+b)(a^2 + b^2 - ab)
    =6 (28-4)=6(24)=144
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