Check my proof: If $a$, $b$, $c$ are distinct, then $a+b+c=0$ if and only if $(a,a^3)$, $(b,b^3)$, $(c,c^3)$ are collinear

Question

Prove that if $a,b,c\in \mathbb{R}$ are all distinct, then $a+b+c=0$ if and only if $(a, a^3), (b,b^3), (c,c^3)$ are collinear.

Let $a,b,c\in \mathbb{R}$ be all distinct. Suppose $a+b+c=0$. Then $c= -(a+b)$. To show that $(a,a^3), (b,b^3),(c,c^3)$ are collinear, it suffices to show that the slope of the line $L_1$ through the points $(a,a^3)$ and $(b,b^3)$ equals the slope of the line $L_2$ through the points $(b,b^3), (c,c^3)$. The slope of $L_1$ is $\dfrac{b^3 - a^3}{b-a} = b^2+ab+a^2$ and the slope of $L_2$ is $c^2+bc+b^2 = (a+b)^2 - b(a+b) + b^2 = b^2+ab+a^2,$ so the slopes are clearly equal.

Now suppose $(a,a^3), (b,b^3),(c,c^3)$ are collinear. Then the slope of the line $L_1$ through $(a,a^3)$ and $(b,b^3)$ must be equal to the slope of the line $L_2$ through $(b,b^3)$ and $(c,c^3)$. Indeed, the slope of a line is the tangent of the angle it makes with the positive $x$-axis, and since $\tan(x+\pi) = \tan(x)$ for all $x\in \mathbb{R}, $ if the slopes are unequal then the angles do not differ from each other by a multiple of $\pi$ and hence the lines cannot be collinear. We then have that $b^2 + ab+a^2 = c^2 + bc+b^2\Rightarrow c^2 - a^2 = b(a-c)\Rightarrow c+a = -b$ because $a,b,c$ are distinct. Thus the result folllows.

Is this proof correct?

Answer 1

The first part of your proof is enough because you can proceed by equivalence

$$\text{aligned points} \ \iff \ \text{equal slopes} \ \iff \ c^2+bc+b^2 = b^2+ab+a^2 \ \iff \ (a-c)(a+b+c)=0 \ \iff \ (a+b+c)=0$$

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Here is an alternate proof:

Using the classical alignment criteria for 3 points $(x_k,y_k)$ for k=1,2,3 which is

$$\begin{vmatrix}x_1&x_2&x_3\\y_1&y_2&y_3\\1&1&1\end{vmatrix}=0$$

This means that we have to show that

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

But this is very easy because the determinant can be factorized in the following way:

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

knowing that $a,b,c$ are all different.

In fact, I just realized that I had already answered a similar question here... with two proofs, this one and another one based on a third degree equation with no term in $x^2$.