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Math Help - cubic formula

  1. #1
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    cubic formula

    Q). Let a,b,c be the roots of x^3 + 2x^2 + 7x + 1 = 0.
    Obtain A^3 +b^3 +C^3

    SOLUTION).
    a + b + c = -2
    ab + bc + ca = 7
    abc = -1
    Now what is (a + b+ c)^3 = a^3 + b^3 + c^3 ++++++++
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by puneet View Post
    Q). Let a,b,c be the roots of x^3 + 2x^2 + 7x + 1 = 0.
    Obtain A^3 +b^3 +C^3

    SOLUTION).
    a + b + c = -2
    ab + bc + ca = 7
    abc = -1
    Now what is (a + b+ c)^3 = a^3 + b^3 + c^3 ++++++++
    I'll do it quite a rough way, I haven't thought of something simpler yet.

    \underbrace{(a+b+c)^3}_{(-2)^3=-8}=\underbrace{a^3+b^3+c^3}_{N}+{\color{red}3ab^2+  3ac^2+3a^2b+3a^2c+3bc^2+3b^2c} \underbrace{+6abc}_{-6}

    N=-2-{\color{red}\text{red part}}

    We'll take care of the red part.

    Factor as much as possible :
    {\color{red}\text{red part}}=3 \bigg(ab(a+b)+ac(a+c)+bc(b+c)\bigg)

    Substitute from a+b+c=-2 :

    \begin{aligned} a+b&=&-2-c \\<br />
a+c&=&-2-b \\<br />
b+c&=&-2-a \end{aligned}

    Thus {\color{red}\text{red part}}=3 \bigg(-2ab-abc-2ac-abc-2bc-abc\bigg)

    {\color{red}\text{red part}}=3 \bigg(-2(ab+ac+bc)-3abc\bigg)=-6*7+9=-33


    Thus N=-2-{\color{red}\text{red part}}=\boxed{31}
    Last edited by Moo; August 23rd 2008 at 03:21 AM. Reason: major mistake
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  3. #3
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    solution

    a^3 +b^3 +c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+bc+ca)
    =(-2)^3 +3(-1) -3(-2)(7)
    =-8-3+42
    =31 ans
    Is this right
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  4. #4
    Moo
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    Quote Originally Posted by puneet View Post
    a^3 +b^3 +c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+bc+ca)
    =(-2)^3 +3(-1) -3(-2)(7)
    =-8-3+42
    =31 ans
    Is this right
    Yup
    Pardon me, I didn't know this factorisation and I made a mistake
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