Factorising a 3 x 3 determinant - What Am I doing Wrong?

Question

$$\begin{vmatrix} 1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \\ \end{vmatrix}$$ subtracting the top row from the middle and bottom rows $$ = \begin{vmatrix} 1 & a & a^3 \\ 0 & (b - a) & (b^3 - a^3)\\ 0 & (c - a) & (c^3 - a^3)\\ \end{vmatrix} $$ expanding the determinant $$ = \begin{vmatrix} (b - a) & (b^3 - a^3)\\ (c - a) & (c^3 - a^3)\\ \end{vmatrix} $$ pulling out factors $$ = (b-a)(c-a) \begin{vmatrix} 1 & (b^2 + a^2)\\ 1 & (c^2 + a^2)\\ \end{vmatrix} $$ and evaluating gives $$ = (b-a)(c-a)(c^2 - b^2) $$ yet the answer in the book gives $$(a+b+c)(a-b)(b-c)(c-a)$$

I am self studying so thanks in advance

Answer 1

Do you know, $$x^3-y^3=(x-y)(x^2+xy+y^2)$$ and not $$x^3-y^3\ne(x-y)(x^2+y^2)$$ in general?

Answer 2

But $b^3-a^3 = (b-a)(b^2+ab+a^2)$ and $c^3-a^3 = (c-a)(c^2+ac+a^2)$. Hence $$(c^2+ac+a^2) - (b^2+ab+a^2) = c^2+ac- b^2-ab = (c-b)(a+b+c).$$

Answer 3

A way to compute the determinant via polynomials, à la Vandermonde.

Consider the polynomial $$ f(x) = \begin{vmatrix} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \\ \end{vmatrix}. $$ Clearly two roots of $f(x)$ are $b$ and $c$, because $f(b)$ is the determinant of a matrix with two equal rows, and similarly for $f(c)$. Also, the coefficient of $x^{3}$ is $$\begin{vmatrix} 1 & b\\ 1 & c \end{vmatrix} = c - b $$ and the coefficient of $x^{2}$ is $0$. Since the coefficient of $x^{2}$ is the negative of the sum of the roots, the three roots of $f(x)$ are $$b, \quad c, \quad - b - c,$$ and thus $$ f(x) = (c - b) (x - b) (x - c) (x + b + c). $$ Now set $x = a$.