Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$

Question

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

My attempt:

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

Where did I go wrong?

Answer 1

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

Answer 2

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

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

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

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

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

Answer 3

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 $K=1$.

Answer 4

Given a $4\times 4$ 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 $$\color{red}{(b-a)(c-a)(c-b)}\color{blue}{(a+b+c)}. $$

Answer 5

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)=$

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

Answer 6

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.

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

(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 :

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

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.