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    the roots of 5x^3 - 16x^2 - 12x + 3 = 0 are alpha, beta and gamma, what is the value of

    alpha^3 + beta^3 + gamma^3?
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    (Original post by fishpaste)
    the roots of 5x^3 - 16x^2 - 12x + 3 = 0 are alpha, beta and gamma, what is the value of

    alpha^3 + beta^3 + gamma^3?

    can it be factorised? ...will try...!
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    (Original post by ogs)
    can it be factorised? ...will try...!
    has to be done using the P5 methods I'm afraid
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    I think it factorised to...

    (5x-1)(x^2+3x-3)
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    (Original post by fishpaste)
    the roots of 5x^3 - 16x^2 - 12x + 3 = 0 are alpha, beta and gamma, what is the value of

    alpha^3 + beta^3 + gamma^3?
    The longhand way:

    Rewrite the eq as x^3 - 16/5 x^2 - 12/5 x + 3/5 = 0.

    Using theorems regarding the roots of equations:

    a+b+c = 16/5
    ab+bc+ac = -12/5
    abc = -3/5.

    Now consider (a+b+c)^3 - you can work out a^2+b^2+c^2 using the first two equations, and simplify to get a^3+b^3+c^3
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    alpha = 1/5
    beta=(3+root21)/2
    gamma=(3-root21)/2

    therefore the sum of the cubes =54.0 3sf

    i don't know how correct it is!
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    The shorthand way is to factorise, although it's not easy to spot the factor
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    a³ + b³ + c³ = (a + b + c)³ - 3(ab + bc + ac)(a + b + c) + 3abc
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    there was an error on the factorising before! .. it really goes to..
    (5x-1)(x^2-3x-3)...
    p5 are you doing edexcel? cos we don't have to do this type of stuff...
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    Oh god, you guys I'm such a twit. I wrote down the wrong equation, the correct equation was

    x^3 - 3x^2 + 5x - 7 = 0

    I'm pretty sure they want me to do it by finding an expression for alpha^3 + beta^3 + gamma^3 in terms of the other things, but I'm not sure how to do this.
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    (Original post by It'sPhil...)
    a³ + b³ + c³ = (a + b + c)³ - 3(ab + bc + ac)(a + b + c) + 3abc
    This is the expression I wanted. Any tips of deriving these kind of expressions in the exam?

    Thanks very much.

    (this is OCR for teh person who asked)
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    From the book you know that

    ax^3 + bx^2 + cx + d=0

    has roots and that

    aS3 + bS2 + cS1=0

    Where S1=alpha+beta+gamma

    and S2=alpha^2 + beta^2 + gamma^2

    etc.
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    IN fact sorry

    aS(3+r) + bS(2+r) + cS(1+r) + dS(r)=0

    and r is an interger.
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    (Original post by PhilC)
    IN fact sorry

    aS(3+r) + bS(2+r) + cS(1+r) + dS(r)=0

    and r is an interger.
    Hm. I don't recognise that at all =o Not in the Cambridge OCR book per chance is it?
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    (Original post by fishpaste)
    This is the expression I wanted. Any tips of deriving these kind of expressions in the exam?

    Thanks very much.

    (this is OCR for teh person who asked)
    Im not sure. I got the expression from a STEP problem I did a while ago. I think the thing to do is expand (a + b + c)^n and see what you get, eg n = 2 is easy...

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

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

    Also always bear in mind you know three things: a + b + c, ab + bc + ac, abc so any expression you derive must be in terms of these only
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    Yes, justr before Miscellenious Exercise 1
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    (Original post by PhilC)
    Yes, justr before Miscellenious Exercise 1
    Ah found it thank you very much!
 
 
 
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