<s>[INST] Factorising a 3 x 3 determinant - What Am I doing Wrong? $$\begin{vmatrix}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3 \\
\end{vmatrix}$$
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
 [/INST] 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 $x = a$.
</s>