<s>[INST] $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$ Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$
Answer: $K=(ab+bc+ca)$
My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3-a^3\end{vmatrix}\\&=\begin{vmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3-a^3\end{vmatrix}\\&=(b-a)(c-a)\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}\\&=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}\\\end{align}$$
I don't know what should I do next, maybe I've made a mistake but I didn't notice it
 [/INST] The mistake is here

$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}$$

These aren't equal to each other. If you want to continue row reduction methods, you should continue with
$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix} = \begin{vmatrix}b-c& c+a\\(b-c) a + b^2 - c^2 &c^2+ca+a^2\end{vmatrix} = (b-c)\begin{vmatrix}1& c+a\\a+b+c &c^2+ca+a^2\end{vmatrix}$$
And you should be able to take it from there.
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