# Equating determinants using properties

Prove without expanding $$\begin{vmatrix} a^3 & a^2 & 1 \\ b^3 & b^2 & 1 \\ c^3 & c^2 & 1 \end{vmatrix} = (ab + bc + ca)\begin{vmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{vmatrix}$$

Well I tried but can't figure out a way to factor out $$(ab + bc + ca)$$ directly without expanding. I can easily show both equal but not directly without expanding.

• Well what i think is to consider if somehow we can convert $(ab + bc + ca)$ into a determinant Apr 2, 2023 at 8:14

$$(ab + bc + ca)\begin{vmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{vmatrix} -\begin{vmatrix} a^3 & a^2 & 1 \\ b^3 & b^2 & 1 \\ c^3 & c^2 & 1 \end{vmatrix}$$ $$=\begin{vmatrix} a^2 & a(ab+bc+ca)+a^3 & 1 \\ b^2 & b(ab+bc+ca)+b^3 & 1 \\ c^2 & c(ab+bc+ca)+c^3 & 1 \end{vmatrix}$$ $$=\begin{vmatrix} a^2 & a^2(a+b+c)+abc & 1 \\ b^2 & b^2(a+b+c)+abc & 1 \\ c^2 & c^2(a+b+c)+abc & 1 \end{vmatrix}=0.$$

• Ah, much neater than mine. +1. Apr 2, 2023 at 8:43
• It's really a beautiful solution 👍 Apr 2, 2023 at 9:02

Let us prove that $$(Xb+Xc+bc)P(X)+Q(X)=0,$$ where $$P(X)=\begin{vmatrix} X^2 &X & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{vmatrix},\quad Q(X)=\begin{vmatrix} X^2 &X^3& 1 \\ b^2 & b^3& 1 \\ c^2 & c^3& 1 \end{vmatrix}.$$

• The well-known Vandermonde determinant $$P(X)$$ can be (re-)computed by writing it $$P(X)=\lambda(X-b)(X-c)$$ where $$bc\lambda=P(0)=bc(b-c)$$ hence (viewing temporarily $$b,c$$ also as polynomial indeterminates) $$\lambda=b-c$$: $$P(X)=(b-c)(X-b)(X-c).$$ Similarly, $$Q(X)=(\mu X+bc(c-b))(X-b)(X-c),$$ and $$\mu=-\begin{vmatrix} b^2 & 1 \\ c^2 &1 \end{vmatrix}=c^2-b^2,$$ i.e. $$Q(X)=((c+b)X+bc)(c-b)(X-b)(X-c),$$ q.e.d.
• Alternatively and more simply (without computing $$P(X),Q(X)$$ explicitely): $$P(X),Q(X)$$ are polynomials of respective degrees $$2,3,$$ and $$Q$$ vanishes at the two (generically distinct) roots $$b,c$$ of $$P$$ hence $$(\alpha X+\beta)P(X)+Q(X)=0,$$ where $$0=\beta P(0)+Q(0)=\beta bc(b-c)+b^2c^2(c-b)$$ i.e. (again by genericity) $$\beta=bc$$ and (identifying the coefficient of $$X^3$$) $$0=\alpha(b-c)+c^2-b^2$$ hence (similarly) $$\alpha=b+c.$$
• Well thanks but can't this question have a more simpler solution 😅 Apr 2, 2023 at 8:35

This is a solution that involves a single expanding but not of one of the original matrices.

Observe that the matrix

$$\begin{vmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{vmatrix}$$

is a Vandermonde matrix with its first and third row swapped.

the determinant of a Vandermonde matrix $$\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}=(b-a)(c-a)(c-b)$$

but if two rows of a matrix are changed, the sign of its determinant also changes.

Therefore we need to show that

$$\begin{vmatrix} a^3 & a^2 & 1 \\ b^3 & b^2 & 1 \\ c^3 & c^2 & 1 \end{vmatrix}=(a-b)(a-c)(b-c)(ab+bc+ca)$$

but $$\begin{vmatrix} a^3 & a^2 & 1 \\ b^3 & b^2 & 1 \\ c^3 & c^2 & 1 \end{vmatrix}=$$

$$\begin{vmatrix} a^3 & a^2 & 1 \\ b^3-a^3 & b^2-a^2 & 0 \\ c^3 & c^2 & 1 \end{vmatrix}=$$

$$(b-a)\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ c^3 & c^2 & 1 \end{vmatrix}=$$

$$(b-a)\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ c^3-a^3 & c^2-a^2 & 0 \end{vmatrix}=$$

$$(b-a)(c-a)\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ c^2+ac+a^2 & c+a & 0 \end{vmatrix}=$$

$$(b-a)(c-a)\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ c^2-b^2+a(c-b) & c-b & 0 \end{vmatrix}=$$

$$(b-a)(c-a)\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ (c-b)(c+b)+a(c-b) & c-b & 0 \end{vmatrix}=$$

$$(b-a)(c-a)(c-b)\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ a+b+c & 1 & 0 \end{vmatrix}=$$

By directly (and easily, given the two zeros) computing the determinant $$\begin{vmatrix} a^3 & a^2 & 1 \\ b^2+ab+a^2 & b+a & 0 \\ a+b+c & 1 & 0 \end{vmatrix}=-(ab+bc+ca)$$

we obtain the result.