Solving equality of $3\times3$ determinant In Basic Mathematics, Lang, exercise of section 17.5...

Gives this answer...

My own calculations gave...
$(x_2-x_1)(x_3^2-x_1^2)-(x_3-x_1)(x_2^2-x_1^2)$
Is it correct?
 A: Personally, I would calculate in a slightly different way, in order to obtain directly a formula easier to memorise:
\begin{align}
\begin{vmatrix}
1&x_1&x_1^2 \\
1&x_2&x_2^2 \\
1&x_3&x_3^2 \\
\end{vmatrix}
&= 
\begin{vmatrix}
0&x_1-x_2&x_1^2-x_2^2 \\
0&x_2-x_3&x_2^2-x_3^2 \\
1&x_3&x_3^2 \\
\end{vmatrix}
=(x_1-x_2)(x_2-x_3)
\begin{vmatrix}
0&1&x_1+x_2 \\
0&1&x_2+x_3 \\
1&x_3&x_3^2 \\
\end{vmatrix} \\
&=(x_1-x_2)(x_2-x_3)\bigl((x_2+x_3)-(x_1+x_2)\bigl) \\
&=\color{red}{(x_1-x_2)(x_2-x_3)(x_3-x_1)}.
\end{align}
A: Yes, because $$(x_2-x_1)(x_3-x_1)(x_3+x_1)-(x_3-x_1)(x_2-x_1)(x_2+x_1)=(x_2-x_1)(x_3-x_1)(x_3-x_2).$$
Another method, which requires basically no calculation: the determinant is a homogeneous cubic in the $x_i$, both fully antisymmetric and invariant under a cyclic permutation of them. It must therefore be proportional to $(x_2-x_1)(x_3-x_2)(x_3-x_1)$, and the proportionality constant is the $x_1^2x_3$ coefficient, which is $\epsilon_{231}=1$.
A: Yes, your answer is  correcr. $$(x_2-x_1)(x_3^2-x_1^2)-(x_3-x_1)(x_2^2-x_1^2)=$$
$$ (x_2-x_1)(x_3-x_1)(x_3+x_1)-(x_3-x_1)(x_2-x_1)(x_2+x_1)$$
Facotor the common terms and simplify to get their answer. 
