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    Hello there!

    I have this really simple looking "show that" question to do but can't seem to verify that LHS = RHS for some reason! Can someone please point out where am I going wrong/guide me on this?



    Here are my workings:





    Basically, I just substituted the value of z in the LHS of the equation to be proven but couldn't manage to arrive at 0.

    Thank you so much!

    Merry Christmas, pals!

    Sid
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    (Original post by Sidhant Shivram)
    Hello there!

    I have this really simple looking "show that" question to do but can't seem to verify that LHS = RHS for some reason! Can someone please point out where am I going wrong/guide me on this?



    Here are my workings:





    Basically, I just substituted the value of z in the LHS of the equation to be proven but couldn't manage to arrive at 0.

    Thank you so much!

    Merry Christmas, pals!

    Sid
    I did your proof/it is not that bad
    I cannot follow your workings after the first expansion

    the plan is
    correctly expand (that is correct in your workings)
    after the cancellations there will be 4 terms
    factorize 3(4+√15)1/3(4-√15)1/3 out of the "bigger" two terms
    then it wraps up after 2 lines
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    (Original post by TeeEm)
    I did your proof/it is not that bad
    I cannot follow your workings after the first expansion

    the plan is
    correctly expand (that is correct in your workings)
    after the cancellations there will be 4 terms
    factorize 3(4+√15)1/3(4-√15)1/3 out of the "bigger" two terms
    then it wraps up after 2 lines
    Ah! I see! Thank you so much! It helped (:
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    (Original post by Sidhant Shivram)
    Ah! I see! Thank you so much! It helped (:
    my pleasure
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    (Original post by Sidhant Shivram)
    ..
    You can get this to fall out very quickly: Note that (4+\sqrt{15})(4-\sqrt{15}) = 1, and so if \alpha = \sqrt[3]{4+\sqrt{15}}, \beta = \sqrt[3]{4-\sqrt{15}} then \alpha\beta = 1 (*)

    Then z^3 = (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha^2\beta + 3 \beta^2\alpha. You can quickly reduce the last two terms using (*).
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    (Original post by DFranklin)
    You can get this to fall out very quickly: Note that (4+\sqrt{15})(4-\sqrt{15}) = 1, and so if \alpha = \sqrt[3]{4+\sqrt{15}}, \beta = \sqrt[3]{4-\sqrt{15}} then \alpha\beta = 1 (*)

    Then z^3 = (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha^2\beta + 3 \beta^2\alpha. You can quickly reduce the last two terms using (*).
    Oh yes! Thanks!
 
 
 
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