When I came across the dilogarithm ladders of Coxeter and Landen, namely,
$$\text{Li}_2\Big(\frac1{\phi^6}\Big)-4\text{Li}_2\Big(\frac1{\phi^3}\Big)-3\text{Li}_2\Big(\frac1{\phi^2}\Big)+6\text{Li}_2\Big(\frac1{\phi}\Big)-\tfrac{7}{5}\zeta(2)=0\tag1$$
$$\text{Li}_2\Big(\frac1{\phi}\Big)+\ln^2\Big(\frac1{\phi}\Big)-\tfrac{3}{5}\zeta(2)=0\tag2$$
where $\phi$ is the golden ratio, I wondered if there was a trilogarithm ladder between cubic algebraic numbers like the tribonacci constant $T$ and transcendental $\zeta(3)$. After some experimentation with Mathematica's integer relations algorithm, I found that the real root of the three equations,
$$T^3=T^2+T+1\tag3$$
$$y^3=y^2+1\tag4$$
$$z^3=z+1\tag5$$
would do. For example,
$$3\text{Li}_3\Big(\frac1{T^4}\Big) -16\text{Li}_3\Big(\frac1{T^3}\Big)-6\text{Li}_3 \Big(\frac1{T^2}\Big)-8\ln^3\Big(\frac1{T}\Big)+12\ln\Big(\frac1{T}\Big)\,\zeta(2)+12\,\zeta(3)=0\;\tag6$$
The other two involved slightly more complicated ladders. A quick search using irreducible cubics failed to find any more.
Q: Is it trivial to find a trilogarithm ladder between $\zeta(3)$ and cubic algebraic numbers, or is there some reason why such a ladder exists for eqns $3,4,5$?
P.S. The real root of $(5)$ is also quite known, being the plastic constant.