determinant linear factorisation

there's this determinant problem I've been working on for several days now whose answer I can't quite get to:

$$D = \left| \begin{array}{ccc} a^3+a^2 & a & 1\\ b^3+b^2 & b & 1\\ c^3+c^2 & c & 1\\ \end{array} \right|$$

Express the determinant as the product of four linear factors.

The given answer is $(a-b)(b-c)(c-a)(a+b+c+1)$ but I'm stuck after getting the first two factors, $(a-b)$ and $(b-c)$:

$$D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & a-b & 0\\ b^2+bc+c^2 & b-c & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$$

whose determinant transposes into:

$$D=(a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & b^2+bc+c^2 & c^3+c^2\\ 1 & 1 & c\\ 0 & 0 & 1\\ \end {array} \right| \\ \Rightarrow D=(a-b)(b-c))1(a^2+ab+b^2)-1(b^2+bc+c^2))\\ =(a-b)(b-c)(a^2+ab+bc+c^2)$$

This is where I am stuck.

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Playing around, and beginning pretty much as you began (except carrying out the row operations correctly): $$D = \begin{vmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b & 0 \\ (b^3-c^3)+(b^2-c^2) & b-c & 0 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b \\ (b^3-c^3)+(b^2-c^2) & b-c \end{vmatrix}\\ = (a-b)(b-c) \begin{vmatrix} a^2+ab+b^2+a+b & 1 \\ b^2+bc+c^2+b+c & 1 \end{vmatrix}\\ = (a-b)(b-c)(a^2+ab+b^2+a+b-b^2-bc-c^2-b-c)\\ = (a-b)(b-c)(a^2+ab+a-bc-c^2-c)\\ = (a-b)(b-c)(a^2+ab+a+ac-ac-bc-c^2-c)\\ = (a-b)(b-c)(a-c)(a+b+c+1),$$ which, annoyingly enough, is minus the answer you're given.

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Indeed, thank you! – Branimir Ćaćić Mar 18 '13 at 8:52

Hint: Sorry, you did a mistake at $D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & a-b & 0\\ b^2+bc+c^2 & b-c & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$. It must be $D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & 1 & 0\\ b^2+bc+c^2 & 1 & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$.

Now, what is the determinant if you expand along the third column? It is a $2\times 2$ determinant $\left| \begin{array}{cc} a^2+ab+b^2 & 1 \\ b^2+bc+c^2 & 1 \\ \end{array} \right|$.

Also, though your approach is fine, there may be alternative ways as well.

Edit: Here is some more hints to find the determinant (possibly "without expanding"):

First note that $D=D_1+D_2$, where $D_1 = \left| \begin{array}{ccc} a^3 & a & 1\\ b^3 & b & 1\\ c^3 & c & 1\\ \end{array} \right|$. Now to find the value of $D_1$, note that if we put $a=b$, or $b=c$, or $c=a$, the determinant vanishes. So, $(a-b)(b-c)(c-a)$ must be factors of the determinant. Now, note that the determinant has a leading term (=product of diagonal entries) $a^3b$. So, the other factor must be a linear...and finally the determinant is actually $-(a-b)(b-c)(c-a)(a+b+c)$. Apply the same technique to find $D_2$ and then add.

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In any case, your transposition step is not correct. How did you get it? – Tapu Mar 18 '13 at 8:06
This also gives the same value when expanded, so I'm not too sure where to start again because I do want those $(a-b)$ and $(b-c)$ factors. – Harris Crescendo Mar 18 '13 at 8:20
Please see my Edits...for another way. – Tapu Mar 18 '13 at 8:26