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Prove that$$beginvmatrix1&1&1\a&b&c\a^3&b^3&c^3endvmatrix=(b-a)(c-b)(c-a)(a+b+c)$$

My attempt:

$$beginalignbeginvmatrix1&1&1\a&b&c\a^3&b^3&c^3endvmatrix&=beginvmatrix0&1&0\a-b&b&c-b\a^3-b^3&b^3&c^3-b^3endvmatrix\&=beginvmatrixc-b&a-b\c^3-b^3&a^3-b^3endvmatrix\&=(c-b)(a-b)beginvmatrix1&1\c^2+cb+b^2&a^2+ab+b^2endvmatrix\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\endalign$$

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
$$
left[beginmatrix
colorred1&colorred1&colorred1&1\colorreda&colorredb&colorredc&d\ a^2&b^2&c^2&colorblued^2\colorreda^3&colorredb^3&colorredc^3&d^3
endmatrixright],
$$ 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
$$beginalign*
&colorred(b-a)(c-a)(c-b)(d-a)(d-b)(d-c)\=&colorred(b-a)(c-a)(c-b)(d^3-colorblue(a+b+c)d^2+cdots).
endalign*$$ Thus we obtain the desired determinant
$$colorred(b-a)(c-a)(c-b)colorblue(a+b+c).
$$

$$beginvmatrix1&1\c^2+cb+b^2&a^2+ab+b^2endvmatrix$$

$$=beginvmatrix1&1-1\c^2+cb+b^2&a^2+ab+b^2-(c^2+cb+b^2)endvmatrix$$

$$=beginvmatrix1&0\c^2+cb+b^2&a^2-c^2+b(a-c)endvmatrix$$

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

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

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 :

$$(a-b)(b-c)(c-a)(a+b+c).tag1$$

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.

$$beginvmatrix1&1&1\x_A&x_B&x_C\y_A&y_B&y_Cendvmatrix=2 times area(ABC)$$

(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 :

$$begincasesy&=&x^3\y&=&ax+bendcases $$

verify

$$x^3-underbrace0_S x^2-ax-b=0.$$

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)=$

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