Symmetry in factor theorem in Determinants

Question

Once my teacher had told me during driving the value of a standard determinant that confused me till now. The value exactly was (a-b)(b-c)(c-a)(a+b+c). Here, (a-b)(b-c)(c-a) had come from factor theorem And he told us that we need a 4 degree term, hence we would need a 4th 1 degree factor symmetric for a,b,c that is (a+b+c) So, **my final request is please clear my concepts on use of symmetry in such type of results as I am very much confused on 2 degree symmetric factors also of the kind k(a²+b²+c²)+k'(ab+bc+ac)

Answer 1

Your determinant is probably

$$D(a,b,c) =\left | \begin{array} \ 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{array}\right |$$

Notice that $D(a,a,c) = 0$ so $D(a,b,c)$ is divisible by $(b-a)$. Similarly it is divisible by $(c-a)$ and $(c-b)$, so by the product $(b-a)(c-a)(c-b)$. Now the quotient $\frac{D(a,b,c)}{(b-a)(c-a)(c-b)}$ is a polynomial, homogeneous of degree $3-2 = 1$. It is also symmetric in $a$, $b$, $c$ ( check that both the numerator and denominator change sign when you flip two of the variables $a$, $b$, $c$). Now a symmetric homogenous of degree $1$ -- it must be the multiple of the sum $a+b+c$. Check that the multiplier is in fact $1$, by looking at the leading terms in $c$ in numerator and denominator.

Note: this can be generalized to $n$ variables.