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    Expand 3x^2/3(2x^2/3+2x^-2/3)

    HELP MEEEE
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    (Original post by stephielola)
    Expand 3x^2/3(2x^2/3+2x^-2/3)

    HELP MEEEE
    First of all, the way you've typed it is ambiguous. Does you mean
     \dfrac{3x^{2}} {3} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})
    Or
     \dfrac{3x^{2}} {3(2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})}

    I suspect the first one, but anyway, what have you tried already?
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    (Original post by KaylaB)
    First of all, the way you've typed it is ambiguous. Does you mean
     \dfrac{3x^{2}} {3} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})
    Or
     \dfrac{3x^{2}} {3(2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})}

    I suspect the first one, but anyway, what have you tried already?
    The first except the root is 2/3 x
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    (Original post by stephielola)
    The first except the root is 2/3 x
    Ah so this?
     3x^{\frac{2} {3}} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})
    Spoiler:
    Show


    This is what I meant about your post being ambiguous :rolleyes:



    So how have you attempted this already?
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    (Original post by KaylaB)
    Ah so this?
     3x^{\frac{2} {3}} (2x^{\frac{2} {3}} + 2x^{-\frac{2} {3}})
    Spoiler:
    Show



    This is what I meant about your post being ambiguous :rolleyes:




    So how have you attempted this already?
    Ahaha!! Yeah sorry about that I have tried multiplying out the bracket using the in-dice law. But I am not sure if I am doing it right
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    (Original post by stephielola)
    Ahaha!! Yeah sorry about that I have tried multiplying out the bracket using the in-dice law. But I am not sure if I am doing it right
    Care to post your working out so I can see if you're heading in the right direction?
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    (Original post by KaylaB)
    Care to post your working out so I can see if you're heading in the right direction?
    I got something like, 6x^2 and an indice of 4/3 +6x^2 and indice to the 0
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    (Original post by stephielola)
    I got something like, 6x^2 and an indice of 4/3 +6x^2 and indice to the 0
    You are correct about the first term :yep:

    With the second term, you are correct in saying that the index would be 0 so it would be  6x^{0} which is the same as what?

    I don't see where you got  6x^{2} from? :hmmmm: The answer should only have 2 terms
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    (Original post by KaylaB)
    You are correct about the first term :yep:

    With the second term, you are correct in saying that the index would be 0 so it would be  6x^{0} which is the same as what?

    I don't see where you got  6x^{2} from? :hmmmm: The answer should only have 2 terms
    Ahh okkkk xx
 
 
 
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