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Further Maths Pure As Exam questions Help! watch

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    Hi I am currently doing practice papers in class and I am stuck of a few questions and wondering if you could help.

    4) Find the values of a^4 + b^4 + c^4 +abc,
    where a, b, c are roots to the equation
    U^3 +16U -9=0

    Thanks
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    (Original post by Alexia_17)
    Hi I am currently doing practice papers in class and I am stuck of a few questions and wondering if you could help.

    4) Find the values of a^4 + b^4 + c^4 +abc,
    where a, b, c are roots to the equation
    U^3 +16U -9=0

    Thanks
    abc should be obvious.

    Then for a^3+b^3+c^3, one way would be to consider the fact that a,b,c all satisfy the cubic, which means we have 3 equations in a,b,c respectively. Adding these together yields a^3+b^3+c^3+16(a+b+c) - 9=0. Note that a+b+c=0 so we just have a^3+b^3+c^3=9

    Now multiply both sides by a+b+c to get (a+b+c)(a^3+b^3+c^3) = 0, hence manipulate the LHS to obtain something that contains only \sum a^4, \sum ab, \sum a and abc
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    (Original post by RDKGames)
    abc should be obvious.

    Then for a^3+b^3+c^3, one way would be to consider the fact that a,b,c all satisfy the cubic, which means we have 3 equations in a,b,c respectively. Adding these together yields a^3+b^3+c^3+16(a+b+c) - 9=0. Note that a+b+c=0 so we just have a^3+b^3+c^3=9

    Now multiply both sides by a+b+c to get (a+b+c)(a^3+b^3+c^3) = 0, hence manipulate the LHS to obtain something that contains only \sum a^4, \sum ab, \sum a and abc
    Hi,
    Thank you for your help. However I am a little confused about how you got to a^3+b^3+c^3+16(a+b+c)+9=0
    Thanks for your help
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    (Original post by Alexia_17)
    Hi,
    Thank you for your help. However I am a little confused about how you got to a^3+b^3+c^3+16(a+b+c)+9=0
    Thanks for your help
    As I mentioned, a,b,c satisfy the cubic. Sub each one into the cubic to get 3 equations. Add the equations.

    I.e. a^3+16a-9=0 and the same for b,c
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    (Original post by RDKGames)
    As I mentioned, a,b,c satisfy the cubic. Sub each one into the cubic to get 3 equations. Add the equations.

    I.e. a^3+16a-9=0 and the same for b,c
    I have currently confused on how I could factorise this:

    ab^3+ac^3+ba^3+bc^3+ca^3+cb^3
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    (Original post by Alexia_17)
    I have currently confused on how I could factorise this:
    ab(b^2+a^2) + bc(b^2+c^2)+ca(c^2+a^2)
    = ab(a^2+b^2+c^2)-abc^2 + bc(a^2+b^2+c^2)-a^2bc + ca(a^2+b^2+c^2)-ab^2c

    Proceed.
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    (Original post by RDKGames)
    ab(b^2+a^2) + bc(b^2+c^2)+ca(c^2+a^2)
    = ab(a^2+b^2+c^2)-abc^2 + bc(a^2+b^2+c^2)-a^2bc + ca(a^2+b^2+c^2)-ab^2c

    Proceed.
    But what would the values be?
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    (Original post by Alexia_17)
    But what would the values be?
    You can figure those out from the original equation!

    Write down the values of ab+bc+ca, a+b+c, and abc by looking at the cubic u^3+16u-9=0.

    What I wrote in the last post can be reduced to simple a case of these three things.
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    (Original post by RDKGames)
    abc should be obvious.

    Then for a^3+b^3+c^3, one way would be to consider the fact that a,b,c all satisfy the cubic, which means we have 3 equations in a,b,c respectively. Adding these together yields a^3+b^3+c^3+16(a+b+c) - 9=0. Note that a+b+c=0 so we just have a^3+b^3+c^3=9

    Now multiply both sides by a+b+c to get (a+b+c)(a^3+b^3+c^3) = 0, hence manipulate the LHS to obtain something that contains only \sum a^4, \sum ab, \sum a and abc
    It looks like they require a^4 + b^4 + c^4 in the question. Or are you giving an example that they can apply to the fourth power sum?
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    (Original post by RDKGames)
    x
    Oh wait I don't think I read the original question or your reply properly
 
 
 
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