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    Not really understanding last part of this Q

    Any help is greatly appreciated Name:  image.jpg
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    You have to use the second part to do the third. What happens if you let x=1/3 in the 2nd expansion you got?
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    As said above. If x=1/3 then you have (from the second part of the question)  (8+3(1/3))^{1/3} = .
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    (Original post by Ano9901whichone)
    You have to use the second part to do the third. What happens if you let x=1/3 in the 2nd expansion you got?
    (Original post by B_9710)
    As said above. If x=1/3 then you have (from the second part of the question)  (8+3(1/3))^{1/3} = .
    okay i see. thanks

    could someone explain why the 'second part' of the question can be used? i.e. the 8+3x inside brackets

    what is not valid about (1+8)^1/3 = 9^1/3 ?
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    (Original post by hi-zen-berg)
    okay i see. thanks

    could someone explain why the 'second part' of the question can be used? i.e. the 8+3x inside brackets

    what is not valid about (1+8)^1/3 = 9^1/3 ?
    I'm sure you've learnt that the binomial expansion of (1+x)^n for non-integer n is only valid for |x| \leq 1. It should be written in your formula booklet as well, next to the expansion.

    Sooo, can't really plug in x=8.

    On the other hand (8 + 3x)^{1/3} is valid for |3x| \leq 1 so x=1/3 is perfectly valid here.
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    (Original post by Zacken)
    I'm sure you've learnt that the binomial expansion of (1+x)^n for non-integer n is only valid for |x| \leq 1. It should be written in your formula booklet as well, next to the expansion.

    Sooo, can't really plug in x=8.

    On the other hand (8 + 3x)^{1/3} is valid for |3x| \leq 1 so x=1/3 is perfectly valid here.
    Ahh I see thanks; have learnt that but it was a bit abstract and didn't care aha
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    (Original post by hi-zen-berg)
    Ahh I see thanks; have learnt that but it was a bit abstract and didn't care aha
    Basically always think about what restrictions apply when it comes to plugging things into expansions!
 
 
 
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