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How to do alpha^4 + Beta^4

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    I know alpha^2+beta^2 and ^3 but i dont know how to expand 4 brackets
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    (Original post by Faheemcg9)
    I know alpha^2+beta^2 and ^3 but i dont know how to expand 4 brackets
    it becomes (a^2+b^2)^2-2a^2b^2, which is the difference of two squares
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    (Original post by MathMoFarah)
    it becomes (a^2+b^2)^2-2a^2b^2, which is the difference of two squares
    (a^2+b^2)^2-2a^2 2b^2 or (a^2+b^2)^2-2a^2 b^2 ? I think you missed the 2 before the beta or maybe im wrong
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    (Original post by Faheemcg9)
    (a^2+b^2)^2-2a^2 2b^2 or (a^2+b^2)^2-2a^2 b^2 ? I think you missed the 2 before the beta or maybe im wrong
    I didn't miss anything
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    (Original post by Faheemcg9)
    I know alpha^2+beta^2 and ^3 but i dont know how to expand 4 brackets
    \displaystyle (\alpha + \beta)^2-2\alpha \beta = \alpha^2 + \beta^2

     \displaystyle \alpha \mapsto \alpha^2 and \displaystyle \beta \mapsto \beta^2

    So \displaystyle \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2-2\alpha^2 \beta^2=[(\alpha+\beta)^2-2\alpha \beta]^2-2\alpha^2 \beta^2

    and so on.

    I wouldn't bother remembering it.
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    (Original post by RDKGames)
    (\alpha + \beta)^2-2\alpha \beta = \alpha^2 + \beta^2

    \alpha \mapsto \alpha^2 and \beta \mapsto \beta^2

    So \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2-2\alpha^2 \beta^2=[(\alpha+\beta)^2-2\alpha \beta]^2-2\alpha \beta

    and so on.

    I wouldn't bother remembering it.
    Thanks
    I'm memorising it because it came up in last years exam
    BTW I see you everywhere on this forum.
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    Well I assume Alpha and a Beta make the average male?
    number to the power of 4 male
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    (Original post by Faheemcg9)
    It came up in last years exam
    Well there you go then, I literally derived it in like 2 lines for you. It's easier to derive it than to remember it.
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    (Original post by MathMoFarah)
    it becomes (a^2+b^2)^2-2a^2b^2, which is the difference of two squares
    Difference of two squares??? Err, no. Unless you want to be bringing \sqrt{2} into it which is a mess.

    (Original post by Faheemcg9)
    Thanks
    I'm memorising it because it came up in last years exam
    BTW I see you everywhere on this forum.
    Well if you are just keep in mind that the last two terms should be squared. I fixed that in my post.
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    (Original post by RDKGames)
    Difference of two squares??? Err, no. Unless you want to be bringing \sqrt{2} into it which is a mess.
    OP wasn't very clear and I thought he meant factorise a^4+b^4 - no other way to do so.
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    (Original post by MathMoFarah)
    OP wasn't very clear and I thought he meant factorise a^4+b^4 - no other way to do so.
    Even then it wouldn't factorise due to a non-integer.
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    (Original post by RDKGames)
    Even then it wouldn't factorise due to a non-integer.
    It still factorises - just not into integers, you can play around with a and b (multiplying by constants, adding/subtracting) to make it integers if that's important for some reason.
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    (Original post by RDKGames)
    Even then it wouldn't factorise due to a non-integer.
    Yes it does? Factorisation doesn't have to be solely integers.
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    (Original post by MathMoFarah)
    It still factorises - just not into integers, you can play around with a and b (multiplying by constants, adding/subtracting) to make it integers if that's important for some reason.
    I meant to say non-rationals, but okay. Just that irrationals in a factorised form are awkward to work with, and I doubt this type of question with irrationals in factorised form would be relevant to anything in FP1's roots of quadratics.
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    (Original post by RDKGames)
    Even then it wouldn't factorise due to a non-integer.
    Yeh it does, that factorisation is very common.


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    (Original post by Faheemcg9)
    I know alpha^2+beta^2 and ^3 but i dont know how to expand 4 brackets
    A more helpful way to handle these might be the following. If
    

\alpha + \beta = A, \quad \alpha \beta = B
    and
    

S_n = \alpha^n + \beta^n
    then
    

S_{n+2} = A S_{n+1} - B S_n
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    And +rep / PRSOM to the first one to derive it (above).
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    (Original post by RichE)
    A more helpful way to handle these might be the following. If
    

\alpha + \beta = A, \quad \alpha \beta = B
    and
    

S_n = \alpha^n + \beta^n
    then
    

S_{n+2} = A S_{n+1} - B S_n
    We could solve this reccurance then in terms of A and B then?


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    (Original post by physicsmaths)
    We could solve this reccurance then in terms of A and B then?


    Posted from TSR Mobile
    Of course. But my point was if you just wanted S4 or S5 then you could quickly use the recurrence to get them without having to figure out any factorisation. *
 
 
 
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