$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, 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

Consider the matrix
 \begin{bmatrix} 1& 1 &1 &1 \\ X & a&b&c \\X^2 & a^2 & b^2 & c^2 \\ X^3 & a^3 & b^3 & c^3 \end{bmatrix}
Its determinant is a Vandermonde determinant, equal to
$$(a-X)(b-X)(c-X)(b-a)(c-a)(c-b)$$
But developping with respect to the first column, you see that the determinant you are looking for is just equal to the coefficient (with a sign minus) of the $X$ term in this polynomial, which is 
$$(-bc-ac-ab)(b-a)(c-a)(c-b)$$
Finally you get that the $K$ you are looking for is equal to
$$K = ab + ac + bc$$
