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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.

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

3 Answers 3

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

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    $\begingroup$ Ah, much neater than mine. +1. $\endgroup$ Apr 2, 2023 at 8:43
  • $\begingroup$ It's really a beautiful solution 👍 $\endgroup$
    – Ash_Blanc
    Apr 2, 2023 at 9:02
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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) $\alpha=b+c.$
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  • $\begingroup$ Well thanks but can't this question have a more simpler solution 😅 $\endgroup$
    – Ash_Blanc
    Apr 2, 2023 at 8:35
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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.

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