Does the polynomial $r^3+s^3+t^3-3rst-1$ have rational roots?

Question

I am looking at natural numbers $n$ of the form

$$n=a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = \det \left(\begin{array}{rrr}a & b & c \\c & a & b \\b & c & a\end{array}\right) $$

with $a,b,c \in \mathbb{Q}$.

Suppose we can find rational numbers $r,s,t$ such that:

$$1 = r^3+s^3+t^3-3rst = (r+s+t)(r^2+s^2+t^2-rs-rt-st) = \det\left(\begin{array}{rrr} r & s & t \\ t & r & s \\ s & t & r \end{array}\right) $$

Then we would have a new "solution" for $n$:

$$n = n\cdot 1 = (a^3+b^3+c^3-3abc)(r^3+s^3+t^3-3rst) = u^3+v^3+w^3-3uvw$$

So the question is if the group:

$$G := \{ (r,s,t) | r,s,t \in \mathbb{Q} , \det(r,s,t) = 1 \}$$

has some interesting points?

Thanks for your help!

Answer 1

I hope, that this it what you are looking for:

$\gamma:=e^{i\frac{2\pi}{3}}$

$A^3+B^3+C^3-3ABC=D^3 \enspace$ with $\enspace D:=(a+b+c)(a+\gamma b+\gamma^2 c)(a+\gamma^2 b+\gamma c)$

and therefore $\enspace D= a^3+b^3+c^3-3abc\,$ .

So, you'll get $\,A,B,C\,$ expressed by $\,a,b,c\,$ which means, that with rational numbers $\,a,b,c\,$ the numbers $\,A,B,C\,$ and $\,\frac{A}{D},\frac{B}{D},\frac{C}{D}\,$ are also rational.