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    I'll post the solution now.
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    (Original post by DomStaff)
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    Have you managed to do it? I finally got it by letting:

    \begin{aligned} f(x) & = a \left( x^3 + \dfrac{bx^2}{a} + \dfrac{cx}{a} + \dfrac{d}{a} \right) = a (x-\alpha_1)(x-\alpha_2)(x-\alpha_3) \end{aligned}

    Comparing the coefficients of the two forms of f(x) gives:

    (1) \qquad 1 = 1

    \begin{aligned} (2) \qquad \dfrac{b}{a} = - \left( \alpha_1 + \alpha_2 + \alpha_3 \right) \ \overset{\alpha_1 = \alpha_2 + \alpha_3 }\implies \ -2\alpha_1 = \dfrac{b}{a} \end{aligned}

    \begin{aligned} (3) \qquad \dfrac{c}{a} = \alpha_1 \left( \alpha_2 + \alpha_3 \right) + \alpha_2 \alpha_3 \ \overset{\alpha_1 = \alpha_2 + \alpha_3 }\implies \ \left( \alpha_1 \right)^2 + \alpha_2 \alpha_3 = \dfrac{c}{a} \end{aligned}

    \begin{aligned} (4) \qquad \dfrac{d}{a} = - \alpha_1  \alpha_2 \alpha_3 \end{aligned}

    The task is to now eliminate \alpha_2 \alpha_3 from the equations, since you already have \alpha_1 in terms of a and b.
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    This literally took forever to type out in Latex! I have put a spoiler on just in case anyone is still working on the problem. If so, your first hint is to write the roots as k, n-k and n.

    There may be a much more elegant way to do this, but my method is certainly correct.

    Spoiler:
    Show
    Let the roots be  k, n-k, n

    Therefore:

     k + n-k + n = 2n = -\frac {b}{a}
    k(n- k) + kn + n(n-k) = n^2 + kn -k^2 = \frac {c}{a}
    k(n- k)n = kn^2 -k^2n= -\frac {d}{a}

    We can rearrange the bottom two equations for  k^2
    n^2 + kn -k^2 = \frac {c}{a} gives us:  k^2 = kn + n^2 -\frac {c}{a}

    kn^2 -k^2n= -\frac {d}{a} gives us:  k^2 = \frac {kn^2 + \frac {d}{a}}{n}=kn + \frac {d}{an}

    Equating these two gives:
     kn + n^2 -\frac {c}{a} = kn + \frac {d}{an}

      n^2 -\frac {c}{a}= \frac {d}{an}

    Now, we can eliminate n:
     n=-\frac {b}{2a}

    Our previous equation then, after some rearranging becomes:
     \frac {b^2 -4ac}{4a^2} = -\frac {2d}{b}

    Multiplying out denominators:
     b^3 -4abc = -8a^2d
     b^3 = 4a(bc-2ad) as desired.


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    (Original post by DomStaff)
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    That's a pretty nice method! What did you think of my one?
    (They seem pretty similar.)
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    (Original post by Khallil)
    That's a pretty nice method! What did you think of my one?

    I personally would say that our methods are similar. As a matter of fact, they are identical once you eliminate your a_2 * a_3 (I just tried your method).

    In terms of length, I'd say they are relatively similar, yours may be slightly shorter. Still, on paper, mine was only half a page.

    Your method is nice. Although, at first I was puzzled as to where your 1=1 came from, but then I realised haha!
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    (Original post by DomStaff)
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    Haha, yep! I tried to make it clear that I didn't get a=1 this time around.
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    i have no idea how people do that latex stuff and write solutions in that neat way.


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    (Original post by physicsmaths)
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    Magic!
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    (Original post by physicsmaths)
    i have no idea how people do that latex stuff and write solutions in that neat way.


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    Did you do the solution in the end?
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    (Original post by DomStaff)
    Did you do the solution in the end?
    yh i did it a didfferent way. i used factorisation etc


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    (Original post by physicsmaths)
    yh i did it a didfferent way. i used factorisation etc


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    Post your solution so we don't have to guess your solution! I always like to see an alternative method.

    You'll be fine with Latex, if you really can't be bothered with it, just write it normally, like (a/b) or b^(3) or whatever.
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    (Original post by DomStaff)
    Post your solution so we don't have to guess your solution! I always like to see an alternative method.

    You'll be fine with Latex, if you really can't be bothered with it, just write it normally, like (a/b) or b^(3) or whatever.
    basically looking at my solution you guys have used standard results where i have had to derive those results. So i have had a hefty amount of working before hand. Then used them results tonprove the rest.


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    basically
    Let the equation be rewritten as an equation with roots alpha, beta , alpha+beta.
    hence it can be written as
    a(x-alpha)(x-beta)(x-(alpha+beta))
    upon expansion prove what alpha and beta must equal i e (-b/a) then its simple from there on. My solution is more along the lones of the other guy. The first solution that was put up.


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    (Original post by physicsmaths)
    basically
    Let the equation be rewritten as an equation with roots alpha, beta , alpha+beta.
    hence it can be written as
    a(x-alpha)(x-beta)(x-(alpha+beta))
    upon expansion prove what alpha and beta must equal i e (-b/a) then its simple from there on. My solution is more along the lones of the other guy. The first solution that was put up.


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    khalil is his name.


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