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    I'm stuck on the question pictured. I can see it's odd cubes - even cubes but that's it? Please help Name:  ImageUploadedByStudent Room1458239013.548104.jpg
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    (Original post by maths_4_life)
    I'm stuck on the question pictured. I can see it's odd cubes - even cubes but that's it? Please help Name:  ImageUploadedByStudent Room1458239013.548104.jpg
Views: 169
Size:  119.7 KB


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    HAve you considered grouping the odd and even cubes together?

    And once you do that, what can you do in terms of factoring out an even term?
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    (Original post by Slowbro93)
    HAve you considered grouping the odd and even cubes together?

    And once you do that, what can you do in terms of factoring out an even term?
    You mean like this....?
    Name:  ImageUploadedByStudent Room1458241075.561199.jpg
Views: 117
Size:  118.9 KB

    Also would they always give you the intermediate step (ie to factor out 16) to help you get to the final answer? How could you just instinctively do that?...not that I've got to that stage yet...


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    (Original post by maths_4_life)
    You mean like this....?
    Name:  ImageUploadedByStudent Room1458241075.561199.jpg
Views: 117
Size:  118.9 KB

    Also would they always give you the intermediate step (ie to factor out 16) to help you get to the final answer? How could you just instinctively do that?...not that I've got to that stage yet...


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    Hey

    Sorry for the delay, when I replied to this I was thinking of the best way of approaching this question and then made a cup of coffee

    So you're correct in terms of grouping the odds and evens so you should get:

     1^3 +3^3 + ... + (n-1)^3 - (2^3 + 4^3 + ... + n^3)

    Now regarding the second bracket, what can you do in terms of factoring terms? (hint: what do 4^3, 6^3, 8^3 all as a common factor and how can the second bracket be written as?)

    In addition, notice you also have positive even terms in the final result. How do you think you can get this? (I'm in my office for a while so take your time )

    To answer your question about the 16, they will most likely give you the expression that they want you to lead towards. In the exam, should you not get to this final answer though, just continue with the rest of the question so you don't lose all of your marks!
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    (Original post by Slowbro93)
    Hey

    Sorry for the delay, when I replied to this I was thinking of the best way of approaching this question and then made a cup of coffee

    So you're correct in terms of grouping the odds and evens so you should get:

     1^3 +3^3 + ... + (n-1)^3 - (2^3 + 4^3 + ... + n^3)

    Now regarding the second bracket, what can you do in terms of factoring terms? (hint: what do 4^3, 6^3, 8^3 all as a common factor and how can the second bracket be written as?)

    In addition, notice you also have positive even terms in the final result. How do you think you can get this? (I'm in my office for a while so take your time )

    To answer your question about the 16, they will most likely give you the expression that they want you to lead towards. In the exam, should you not get to this final answer though, just continue with the rest of the question so you don't lose all of your marks!
    Thank you for your reply! I can see that all those numbers are factors of 16 but I don't know how I can write it like that?


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    (Original post by Slowbro93)
    Hey

    Sorry for the delay, when I replied to this I was thinking of the best way of approaching this question and then made a cup of coffee

    So you're correct in terms of grouping the odds and evens so you should get:

     1^3 +3^3 + ... + (n-1)^3 - (2^3 + 4^3 + ... + n^3)

    Now regarding the second bracket, what can you do in terms of factoring terms? (hint: what do 4^3, 6^3, 8^3 all as a common factor and how can the second bracket be written as?)

    In addition, notice you also have positive even terms in the final result. How do you think you can get this? (I'm in my office for a while so take your time )

    To answer your question about the 16, they will most likely give you the expression that they want you to lead towards. In the exam, should you not get to this final answer though, just continue with the rest of the question so you don't lose all of your marks!
    Name:  ImageUploadedByStudent Room1458244938.638013.jpg
Views: 112
Size:  125.6 KBwhat about this?


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    (Original post by maths_4_life)
    Name:  ImageUploadedByStudent Room1458244938.638013.jpg
Views: 112
Size:  125.6 KBwhat about this?


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    Check your negative and positive signs, do they equal the original expression?
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    (Original post by maths_4_life)
    Name:  ImageUploadedByStudent Room1458244938.638013.jpg
Views: 112
Size:  125.6 KBwhat about this?


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    Remember that if you have -(2^3 + 4^3 + 6^3 + \cdots + n^3) = -((1\times 2)^3 + (2 \times 2)^3 + (3\times 2)^2 + \cdots +(\frac{n}{2} \times 2)^3 = -2^3\left(1^3 + 2^3 + \cdots + \left(\frac{n}{2}\right)^3 \right)
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    (Original post by Slowbro93)
    Hey

    Sorry for the delay, when I replied to this I was thinking of the best way of approaching this question and then made a cup of coffee

    So you're correct in terms of grouping the odds and evens so you should get:

     1^3 +3^3 + ... + (n-1)^3 - (2^3 + 4^3 + ... + n^3)

    Now regarding the second bracket, what can you do in terms of factoring terms? (hint: what do 4^3, 6^3, 8^3 all as a common factor and how can the second bracket be written as?)

    In addition, notice you also have positive even terms in the final result. How do you think you can get this? (I'm in my office for a while so take your time )

    To answer your question about the 16, they will most likely give you the expression that they want you to lead towards. In the exam, should you not get to this final answer though, just continue with the rest of the question so you don't lose all of your marks!
    I mean this...Name:  ImageUploadedByStudent Room1458245137.576594.jpg
Views: 121
Size:  125.7 KB


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    (Original post by maths_4_life)
    I mean this...Name:  ImageUploadedByStudent Room1458245137.576594.jpg
Views: 121
Size:  125.7 KB


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    That is correct Now how would you get the positive even cube terms? (consider adding something to the expression)
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    (Original post by Slowbro93)
    That is correct Now how would you get the positive even cube terms? (consider adding something to the expression)
    I thought I already had the positive even cube terms.... What do I have at the moment? I'm confused sorry!



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    (Original post by maths_4_life)
    I thought I already had the positive even cube terms.... What do I have at the moment? I'm confused sorry!



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    So what you have is correct, but the expression you have is: 1^3 + 3^3 + ... + (n-1)^3, whereas you want to have as the final bit: 1^3 + 2^3 + 3^3 ... + n^3

    What would you need to do to get that?
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    (Original post by Slowbro93)
    So what you have is correct, but the expression you have is: 1^3 + 3^3 + ... + (n-1)^3, whereas you want to have as the final bit: 1^3 + 2^3 + 3^3 ... + n^3

    What would you need to do to get that?
    I really don't know :/


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    (Original post by maths_4_life)
    I really don't know :/


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    COnsider adding a new expression (Hint )
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    (Original post by maths_4_life)
    I really don't know :/


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    Surely if I add something to my answer them the result will mean that I no longer have an expression equal to 1^3 - 2^3....-n^3 which I was given to start with? I'm sure I'm wrong though


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    (Original post by maths_4_life)
    Surely if I add something to my answer them the result will mean that I no longer have an expression equal to 1^3 - 2^3....-n^3 which I was given to start with? I'm sure I'm wrong though


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    Yes, but how would you make it equal?

    For example, if I had 1+2+3, this is also the same as 1+2+3+ 4 - 4
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    (Original post by Slowbro93)
    So what you have is correct, but the expression you have is: 1^3 + 3^3 + ... + (n-1)^3, whereas you want to have as the final bit: 1^3 + 2^3 + 3^3 ... + n^3

    What would you need to do to get that?
    Ohhhh maybe I do get it.... I add the 2^3 ....n^3 to the part on the left and then I counteract this by multiplying the part in the brackets by 2 so overall nothing is added....its just rearranged?


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    (Original post by maths_4_life)
    Ohhhh maybe I do get it.... I add the 2^3 ....n^3 to the part on the left and then I counteract this by multiplying the part in the brackets by 2 so overall nothing is added....its just rearranged?


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    Yes!
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    (Original post by maths_4_life)
    Ohhhh maybe I do get it.... I add the 2^3 ....n^3 to the part on the left and then I counteract this by multiplying the part in the brackets by 2 so overall nothing is added....its just rearranged?


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    Correct
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    (Original post by Zacken)
    Yes!
    Thank you sooooo much!!!


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