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    Using Pascal's Triangle only, and not binomial expansion, how would you work this question out:

    [ 1 + (x - 1)^2]^3

    This is what I have been doing so far, and I know I'm definitely making a mistake because the answer I get is different from the answer in the back of my text book.

    My working out is as follows:

    [1 + (x - 1)^2]^3

    1 + (x - 1)^6

    1 + [-1(1 - x)]^6

    1 + (1 - x)^6 This part is wrong I think, but I need help here.

    1 + (1 + 6(-x) + 15(-x)^2 + 20(-x)^3 + 15(-x)^4 + 6(-x)^5 + 1(-x)^6]

    1 + 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6

    2 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6


    The actual answer is: 8 - 24x + 36x^2 - 32x^3 + 18x^4 - 6x^5 + x^6


    Sorry about the length of this post but I'm getting really :mad: about this!!!
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    I don't agree with your first step.
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    (Original post by Narik)
    Using Pascal's Triangle only, and not binomial expansion, how would you work this question out:

    [ 1 + (x - 1)^2]^3

    This is what I have been doing so far, and I know I'm definitely making a mistake because the answer I get is different from the answer in the back of my text book.

    My working out is as follows:

    [1 + (x - 1)^2]^3

    1 + (x - 1)^6

    1 + [-1(1 - x)]^6

    1 + (1 - x)^6 This part is wrong I think, but I need help here.

    1 + (1 + 6(-x) + 15(-x)^2 + 20(-x)^3 + 15(-x)^4 + 6(-x)^5 + 1(-x)^6]

    1 + 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6

    2 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6


    The actual answer is: 8 - 24x + 36x^2 - 32x^3 + 18x^4 - 6x^5 + x^6


    Sorry about the length of this post but I'm getting really :mad: about this!!!
    Put it in latex please?
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    (Original post by Narik)
    My working out is as follows:

    [1 + (x - 1)^2]^3

    1 + (x - 1)^6
    (a+b)^3=a^3 + 3 a^2 b + 3 a b^2 + b^3

    Do you see the 1 3 3 1 ?

    rather than a+b^3 which is what you seem to have done. (or even a^3 + b^3)
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    (Original post by Narik)
    Using Pascal's Triangle only, and not binomial expansion, how would you work this question out:

    [ 1 + (x - 1)^2]^3

    This is what I have been doing so far, and I know I'm definitely making a mistake because the answer I get is different from the answer in the back of my text book.

    Pascals:

    ...1
    ..1 1
    .1 2 1
    1 3 3 1 Thus use this line.
    My working out is as follows (AMENDED):

    [1 + (x - 1)^2]^3



[1 + x^2-2x+1]^3



[x^2-2x+2]^3



[x^2-(2x-2)]^3



1(x^2)^3 - 3(x^2)^2(2x-2)^1 + 3(x^2)^1(2x-2)^2 - 1(2x-2)^3



x^6 - 3x^4(2x-2) + 3x^2(4x^2-8x+4) - (8x^3-24x^2+24x-8)



x^6 - 6x^5 + 6x^4 + 12x^4 - 24x^3 + 12x^2 - 8x^3 +24x^2 - 24x +8



x^6 - 6x^5 + 18x^4 -32x^3 + 36x^2 - 24x + 8
    [/thread]
    • Wiki Support Team
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    Wiki Support Team
    (Original post by Small123)
    [/thread]
    Read the posting guidelines, please. Full solutions tend to be frowned upon. Very nice of you, but not necessarily the most helpful thing to do.
 
 
 
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