Solving equality of $3\times3$ determinant

Question

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?

Answer 1

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}

Answer 2

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 $\epsilon_{231}=1$.

Answer 3

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.