Factorizing the determinant Here, in this question, we have to factorize this determinant. I understood the (a-b), (b-c), (c-a) part but I am not able to understand

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*how to find if determinant is symmetrical in a, b, c.

*I know that rest factor will be of degree 2 as we have got (a-b), (b-c), (c-a), but as here (a^2 + b^2 + c^2) and (ab + bc + ca) are possible factors of degree 2, can't be (a + b + c)^2 will be the 2nd degree factor?

Book HK Das Engineering Mathematics page 388 example no. 36
Book HK Das Engineering Mathematics page 388 example no. 36

(1) If you put $a=b,b=c$ or $c=a$ then, $\Delta=0$.
Being symmetric about $a,b,c$ means that if $(a_0,b_0,c_0)$ is a solution to $\Delta=0$, then $(b_0,a_0,c_0)$ is also a solution and so are all other permutations of $(a_0,b_0,c_0)$. This can be seen easily. For now, we verify it for the $3$ permutations $(a_0,b_0,c_0),(b_0,a_0,c_0),(c_0,b_0,a_0)$:
$$\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=-\begin{vmatrix}1 & 1 & 1\\b_0^2&a_0^2&c_0^2\\b_0^3&a_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\c_0^2&b_0^2&a_0^2\\c_0^3&b_0^3&a_0^3\end{vmatrix}\mathrm{~~and~~}\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=0$$
$$\Rightarrow\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\b_0^2&a_0^2&c_0^2\\b_0^3&a_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\c_0^2&b_0^2&a_0^2\\c_0^3&b_0^3&a_0^3\end{vmatrix}=0$$
$\therefore a=a_0,b=b_0,c=c_0$ and $a=b_0,b=a_0,c=c_0$ and $a=c_0,b=b_0,c=a_0$ are all solutions to $\Delta=0$.

(2) $(a+b+c)^2$ may or may not be a possible factor. Notice that if it is then
$$(a+b+c)^2=1\cdot(a^2+b^2+c^2)+2\cdot(ab+bc+ca).$$
The point of taking $(a^2+b^2+c^2)$ and $(ab+bc+ca)$ is that any $2$ degree factor in terms of $a,b,c$ can be written in the form
$$k\cdot(a^2+b^2+c^2)+l\cdot(ab+bc+ca)$$
whereas, there are possible $2$ degree factors of $\Delta=0$ that may not be multiples of $(a+b+c)^2$.
