Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}


*I tried to zero some elements and expand until I reach the Right hand side.

*Also tried C1-C3, C2-C3 then decompose the determinant into two determinants and taking common factors. But I couldn't get (ab + ac + bc) part.

*I can only use the properties shown here http://www.vitutor.com/alg/determinants/properties_determinants.html

First note that the determinant is cyclic. Hence, it is of the form $f(a,b,c)$, where $f$ is a polynomial of degree $5$. Further, we have
$f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$, which means $(a-b)$, $(b-c)$ and $(c-a)$ are factors, i.e., the determinant is $g(a,b,c)(a-b)(b-c)(c-a)$, where $g(a,b,c)$ is a cyclic polynomial of degree $2$. Any cyclic polynomial of degree $2$ is of the form $x(a^2+b^2+c^2)+y(ab+bc+ca)$. Setting $a=0$, we see that the determinant is $b^2c^3-b^3c^2 = b^2c^2(c-b)$. We also have
$$f(0,b,c) = g(0,b,c)(-b)(b-c)c = \left(x(b^2+c^2)+ybc \right)bc(c-b)$$
This means we need
$$\left(x(b^2+c^2)+ybc \right)bc(c-b) = b^2c^2(c-b) \implies x(b^2+c^2)+ybc = bc \implies x=0,y=1$$
Hence, we obtain that the determinant is
$$(ab+bc+ca)(a-b)(b-c)(c-a)$$
