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Uchiha Sasuke

24/09/2017 at 14:33
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\(a+b+c=0\) and \(abc=11\). Calculate \(a^3+b^3+c^3\).



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  • ...
    Kaya Renger Coordinator 24/09/2017 at 14:46

    Need to prove :<

    a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

    We have :

    a3 + b3 + c3 - 3abc 

    = (a + b)3 - 3ab(a + b) + c3 - 3abc

    = [(a + b)3 + c3] - 3ab(a + b + c)

    = (a + b + c)[(a + b)2 - (a + b).c + c2]- 3ab(a + b + c)

    = (a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab)

    = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) 

    So :

    a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

    But a + b + c = 0

    => a3 + b3 + c3 - 3abc = 0

    => a3 + b3 + c3 = 3abc 

    => a3 + b3 + c3 = 3.11 = 33

    So ........ 

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  • ...
    Faded 28/01/2018 at 20:41

    Need to prove :<

    a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

    We have :

    a3 + b3 + c3 - 3abc 

    = (a + b)3 - 3ab(a + b) + c3 - 3abc

    = [(a + b)3 + c3] - 3ab(a + b + c)

    = (a + b + c)[(a + b)2 - (a + b).c + c2]- 3ab(a + b + c)

    = (a + b + c)(a2 + 2ab + b2 - ac - bc + c2 - 3ab)

    = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) 

    So :

    a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

    But a + b + c = 0

    => a3 + b3 + c3 - 3abc = 0

    => a3 + b3 + c3 = 3abc 

    => a3 + b3 + c3 = 3.11 = 33

    So ........



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