Dtjytyk

Prove that$$begin{vmatrix}1&1&1\a&b&c\a^3&b^3&c^3end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$

My attempt:

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

Where did I go wrong?

Note that $(a^2+ab)-(c^2+cb)=(a-c)(a+c)+b(a-c)=(a-c)(a+b+c)$

Given a $4times 4$ Vandermonde matrix
$$
left[begin{matrix}
color{red}1&color{red}1&color{red}1&1\color{red}a&color{red}b&color{red}c&d\ a^2&b^2&c^2&color{blue}{d^2}\color{red}{a^3}&color{red}{b^3}&color{red}{c^3}&d^3
end{matrix}right],
$$ note that the desired determinant is the minor of $d^2$. Considering Laplace expansion, it appears as a minus(-) coefficient of $d^2$ in the Vandermonde determinant
$$begin{align*}
&color{red}{(b-a)(c-a)(c-b)}(d-a)(d-b)(d-c)\=&color{red}{(b-a)(c-a)(c-b)}(d^3-color{blue}{(a+b+c)}d^2+cdots).
end{align*}$$ Thus we obtain the desired determinant
$$color{red}{(b-a)(c-a)(c-b)}color{blue}{(a+b+c)}.
$$

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

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

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

$$=a^2-c^2+b(a-c)$$

$$=(a-c)(a+c+b)$$

You can get it immediately:
$$sum_{cyc}(a^3c-a^3b)=(a+b+c)(a-b)(b-c)(c-a)$$
because for $a=b$ or $a=c$ or $b=c$ our determinant is equal to zero, which gives
$$K(a+b+c)(a-b)(b-c)(c-a)$$ and it's enough to check the coefficient before $a^3c$, which is $1$, which gives $K=1$.

Here is a geometrical interpretation of the formula :

$$(a-b)(b-c)(c-a)(a+b+c).tag{1}$$

I use the term "interpretation" because I have not written here a complete proof but rather an inductive way to obtain (1).

Indeed, if the presence of factors $(a-b), (b-c), (c-a)$ look very natural (the determinant is zero if at has 2 identical columns), this is apparently not the case for factor $(a+b+c)$.

Here is a context giving a natural interpretation to this factor.

Reminder (see Theorem in https://proofwiki.org/wiki/Area_of_Triangle_in_Determinant_Form ) : Let $A,B,C$ be $3$ points in the plane.

$$begin{vmatrix}1&1&1\x_A&x_B&x_C\y_A&y_B&y_Cend{vmatrix}=2 times area(ABC)$$

(being understood that, is $A,B,C$ are all different, this determinant is zero iff $A,B,C$ are aligned).

Here, we take points $A,B,C$ on the curve $Gamma$ with equation $y=x^3$. If they are distinct, they are aligned iff their abscissas are such that :

$$x_A+x_B+x_C=0$$

which is rather convincing when one looks at curve $Gamma$ (Fig. 1) :

Fig. 1 : the sum of abscissas of aligned points $A,B,C$ is $-1.5+0.5+1=0$.

Now, the proof : the abscissas of intersection points of :

$$begin{cases}y&=&x^3\y&=&ax+bend{cases} $$

verify

$$x^3-underbrace{0}_S x^2-ax-b=0.$$

The sum of roots $S$ (coefficient of $-x^2$ according to Viète's formulas) is thus zero.

1) By inspection : The determinant $= 0$ for $a=b$; $a=c$, and $b=c$ (Why?).

Hence $pm (c-b)$, $pm(a-b)$, and $pm (a-c)$ are factors.

You got $(c-b)(a-b)[a^2+ab -(c^2+cb)]$.

Hence $pm (a-c)$ is a factor of $[a^2+ab -(c^2+cb)]$.

$a^2-c^2+b(a-c)=$

$ (a-c)(a+c)+b(a-c)=$

$(a-c)(a+b+c).$