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    I know the rules for manipulating determinants but I can't seem to factorise this:

    \begin{vmatrix} 1 & a & a^3 \\1 & b & b^3 \\1 & c & c^3 \end{vmatrix}

    I can see +/-(a-b) and by symmetry b-c and a-c are factors but i can't manipulate the determinant to do this which is required. Any help?
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    Well what have you got as the determinant?
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    You should find it is:

      \mathrm{det} = bc^3 + b^3c -a(c^3-b^3) + a^3(c-b)

     = bc(c^2-b^2) -a(c-b)(c^2+bc+b^2) + a^3(c-b)

    And after seeing a 'difference of two squares' on the first term, the 'c-b' factor should be obvious. Try similar methods to show the factors of 'b-a' and 'c-a'.
    Open after trying
     \therefore \mathrm{det} = (c-b)(bc^2 - ac^2 + a^3 -ab^2 + b^2c -abc) [just a slight rearrangement of the terms to make things a little clearer]

     = (c-b)(c^2(b-a) - a(b^2-a^2) +bc(b-a))

    And again, after seeing the 'difference of two squares' on the second term, the 'b-a' factor appears.
    Open after trying
     \therefore \mathrm{det} = (c-b)(b-a)(c^2 -a^2 +bc -ab) [another slight rearrangement of the terms]

     = (c-b)(b-a)((c+a)(c-a) + b(c-a))

    And I'm sure you must be able to take it from there.
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    Thanks; I was hoping I wouldn't have to go through that approach though.

    Can there possibly be a way of taking the factors from the determinant itself. I can do this with other determinants; just not this one. See post #8 on the link below:

    http://www.thestudentroom.co.uk/showthread.php?t=451159
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    If you have an MEI FP2 textbook, look at p89.
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    No sorry- I only have the AQA FP4 textbook http://store.aqa.org.uk/qual/pdf/AQA-MFP4-TEXTBOOK.PDF
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    OK; (a-b)(b-c)(c-a) are factors by symmetry you have.

    What degree is polynomial you seek? Consider any term in the expansion of determinant.

    Cycling a,b and c leave determinant unchanged, and also leave product of three already-found factors unchanged. So a+b+c must be remaining factor. Now we're left with possible constant. Check by considering coefficient of one term and you should be left with +/- 1 depending on how you write the first three factors.
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    (Original post by Palabras)
    Thanks; I was hoping I wouldn't have to go through that approach though.

    Can there possibly be a way of taking the factors from the determinant itself. I can do this with other determinants; just not this one. See post #8 on the link below:

    http://www.thestudentroom.co.uk/showthread.php?t=451159
    If, instead of column operations, like in that link, row operations are allowed to be performed similarly, then in an almost identical way, but with rows rather than columns, subtract the first row from the second and the third row, and then take out appropriate factors of the second and third row. Then compute the resulting discriminant.
 
 
 
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