The Lucas and Fibonacci numbers solve the Pell equation,
$$L_n^2-5F_n^2=4(-1)^n\tag1$$
The tribonacci numbers $z = T_n$ are positive integer solutions to the cubic Pell-type equation,
$$27 x^3 - 36 x y^2 + 38 y^3 - 342 x y z + 96 y^2 z - 144 x z^2 + 456 y z^2 + 1316 z^3 = \color{brown}{3^6}\tag2$$
where $x,y$ are associated sequences to the tribonacci. Or more simply,
$$a^3 - 2 a^2 b + 2 b^3 - a^2 c - 2 a b c + 2 b^2 c + a c^2 + 2 b c^2 + c^3=1\tag3$$
where $a,\;b,\;c = T_{n-1},\;T_{n-2},\;T_{n-3}$. See this post.
Question: For aesthetics, I wish to simplify. Is it possible to reduce $3$rd deg eqns $F(x,y,z)=0$ like $(2)$ and $(3)$ to the simpler form without cross-terms,
$$a_1p^3+a_2q^3+a_3r^3+a_4pqr+a_5 = 0\tag4$$
for integer $a_i$ using a transformation with rational coefficients?
Note: For example, I tried,
$$x = c_1p+c_1q+c_3r$$
$$y = c_4p+c_5q+c_6r$$
$$z = c_7p+c_8q+c_9r$$
Substituting it into $(2)$ and getting rid of the cross-terms entailed a system of six equations. For simplicity, I assumed three of the $c_i = 1$. It was solvable, but my final equation was a quadratic, hence not all the $c_i$ were rational. However, perhaps I was not using the right assumptions.
So is there a general transformation with rational coefficients to get the form of $(4)$, or does it happen only for particular cubics $F(x,y,z)=0$?