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    I have this question.

     x(2x^-\frac{1}{3})^4

    Now, after a few days practice, and starting on negative and fractional indices yesterday, I believe I understand indices to the extent that I can implement the techniques required to solve this individually. What I am unsure about is the order I need to implement them in. Should I be expanding the brackets first i.e  (2x^\frac{2}{3})^4 and then  16x^\frac{8}{3} , then  \sqrt [3] 16x^8 or should I be doing this later, and doing other things first? Or is the order irrelevant? It seems the order matters because I know what the solution to this question is, but I can't quite seem to reach it.
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    Brackets first, remember ur BODMAS (order of operations)
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    So I need to do everything inside the brackets before I expand them? E.g  x(\frac{1} {2x^\frac{1}{3}})^4 = x(\frac{1}{\sqrt[3] 2x})^4

    If this is the case, would I then expand them and then multiply by the  ^4?
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    one way you could do this would to do the brackets 4 times, as it's to the power of 4, you're not multiplying the brackets by 4

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    i think
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    (Original post by Illidan2)
    So I need to do everything inside the brackets before I expand them? E.g  x(\frac{1} {2x^\frac{1}{3}})^4 = x(\frac{1}{\sqrt[3] 2x})^4

    If this is the case, would I then expand them and then multiply by the  ^4?
    The cube root only 'acts' on x - not the 2 as well.
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    (Original post by minertommy)
    one way you could do this would to do the brackets 4 times, as it's to the power of 4, you're not multiplying the brackets by 4

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    i think
    I know, what I meant was to multiply the contents of the brackets by the power four, I.e the contents of the brackets by themselves four times, perhaps it was a poor choice of words on my part.
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    (Original post by Muttley79)
    The cube root only 'acts' on x - not the 2 as well.
    Oh I see, so 2 takes the place of 1, and x remains in the square root alone?
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    (Original post by Illidan2)
    Oh I see, so 2 takes the place of 1, and x remains in the square root alone?
    The 2 should be on the numerator.
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    So then I have  x(\frac{16}{\sqrt[3]x^4}) and then  \frac{16x}{\sqrt[3]x^5} . Where do I go from there? I'm not sure if what i've done is right, because I can't see how I get to the solution from this.

    Edit: Unless  x(\frac{16}{\sqrt[3]x^4}) = \frac{16} {\sqrt[3]x^4}x ? In hindsight, I think it is this.
 
 
 
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