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    jgfjgjghh
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    (3x-1)(3x-1)(3x-1)=[(3x-1)(3x-1)](3x-1)

    simply ignore the final bracket until you've worked out the ones in the [] brackets, then using the same method mulitply out what you get from [] with the (3x-1) on the end

    EDIT: fatchicksftw.
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    (3x-1)(3x-1)(3x-1)=(3x-1)(9x^2-6x+1)
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    9x^2 -6x + 1
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    Binomial expansion??
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    a)Try doing just simply (3x-1)(3x-1) and what do you get?

    b)Then you do (3x-1)(your answer from part a)
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    sorry, did i just MERK YOU ALL!

    muhahahaha.
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    steve2005 laid out the way i would explain doing it OP
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    If you know binomial expansion, then it's just (3x-1)^3
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    (Original post by steve2005)
    (3x-1)(3x-1)(3x-1)=(3x-1)(9x^2-6x+1)
    I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
    Could you please explain
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    Expanding the second two brackets is just like expanding a quadratic function (ax+b)(cx+d); then you just multiply through again.

    Say you have something with 5 brackets (...)(...)(...)(...)(...), then you can tackle it by multiplying everything in the 4th bracket with everything in the 5th, and then everything in the 3rd with what you just worked out for the 4th and 5th, then everything in the 2nd with everything that you just worked out for the 3rd, 4th and 5th, and so on... for example:

    \newline (a+b)(c+d)(e+f)(g+h)(i+j)\newlin  e

= (a+b)(c+d)(e+f)(gi + gj + hi + hj)\newline

= (a+b)(c+d)(egi + egj + ehi + ehj + fgi + fgj + fhi + fhj)\newline

= (a+b)(cegi + cegj + cehi + cehj + cfgi + cfgj + cfhi + cfhj + degi + degj + dehi + dehj + dfgi + dfgj + dfhi + dfhj)\newline

= acegi + acegj + acehi + acehj + acfgi + acfgj + acfhi + acfhj + adegi + adegj + adehi + adehj + adfgi + adfgj + adfhi + adfhj + bcegi + bcegj + bcehi + bcehj + bcfgi + bcfgj + bcfhi + bcfhj + bdegi + bdegj + bdehi + bdehj + bdfgi + bdfgj + bdfhi + bdfhj

    Simples *squeak*

    But really, expanding brackets is quite basic GCSE stuff... what level are you working at here?
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    (Original post by ilovefatchicks)
    I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
    Could you please explain
    (a + b)^2 = (a + b)(a + b) = a(a+b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2

    Apply this to your brackets.
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    OMG a maths question on tsr i can actually do!!!

    So,

    (3x-1)(3x-1)(3x-1)
    = (3x-1)(9x^2 -3x - 3x +1)
    = 27x^3 - 9x^2 -9x^2 + 3x
    -9x^2 +3x +3x -1
    = 27x^3 - 27x^2 + 9x -1

    (probably beaten to it but heyho..)

    EDIT: why does it say -2 at the end?!? Sorry if i'm wrong!!
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    (Original post by ilovefatchicks)
    I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
    Could you please explain
    Thats because you multiply 3x by -1 to get -3x, and then again. You then add the answers together like so: -3x + -3x = -6x

    Remember to do that as well as 3x*3x and -1*-1.
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    (3x-1) * (3x-1)

    FOIL
    First = 3x * 3x = 9x^2
    Outside = -1*-1 = +1
    inside = -1 * 3x = -3x
    Last = -1 * 3x = -3x
    collect terms

    9x^2-6x+1
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    (Original post by nuodai)
    Expanding the second two brackets is just like expanding a quadratic function (ax+b)(cx+d); then you just multiply through again.

    Say you have something with 5 brackets (...)(...)(...)(...)(...), then you can tackle it by multiplying everything in the 4th bracket with everything in the 5th, and then everything in the 3rd with what you just worked out for the 4th and 5th, then everything in the 2nd with everything that you just worked out for the 3rd, 4th and 5th, and so on... for example:

    \newline (a+b)(c+d)(e+f)(g+h)(i+j)\newlin  e

= (a+b)(c+d)(e+f)(gi + gj + hi + hj)\newline

= (a+b)(c+d)(egi + egj + ehi + ehj + fgi + fgj + fhi + fhj)\newline

= (a+b)(cegi + cegj + cehi + cehj + cfgi + cfgj + cfhi + cfhj + degi + degj\newline \qquad + dehi + dehj + dfgi + dfgj + dfhi + dfhj)\newline

= acegi + acegj + acehi + acehj + acfgi + acfgj + acfhi + acfhj + adegi + adegj + adehi + adehj + adfgi + adfgj + adfhi + adfhj + bcegi + bcegj + bcehi + bcehj + bcfgi + bcfgj + bcfhi + bcfhj + bdegi + bdegj + bdehi + bdehj + bdfgi + bdfgj + bdfhi + bdfhj

    Simples *squeak*
    Compare the meerkat dot com, compare the market dot com.
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    (Original post by ilovefatchicks)
    I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
    Could you please explain
    (a+b)^2=a^2+2ab+b^2

    (5a+3b)^2=25a^2+30ab+9b^2

    (4a+3b)^2=16a^2+24ab+9b^2

    etc
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    (Original post by PGtips92)
    EDIT: why does it say -2 at the end?!? Sorry if i'm wrong!!
    You're correct, it's -1.
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    e=mc2 :ninja:
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    (Original post by Arctic_wombaT)
    e=mc2 :ninja:
    Wow, I'm completely taken aback by your wit... you should consider a career in stand-up comedy
 
 
 

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