Tribonacci Treat

Let $\space x = \displaystyle \lim_{n\rightarrow \infty} {\dfrac {F_{n+1}}{F_n} } = \lim_{n\rightarrow \infty} {\dfrac {F_n+F_{n-1}} {F_n} } = 1+\dfrac{1}{x} \\ \Rightarrow x^2 -x -1 = 0\quad \Rightarrow x = \dfrac {A+\sqrt{B}}{C} = \dfrac {1+\sqrt{5}}{2}$

Let $\space y = \displaystyle \lim_{n\rightarrow \infty} {\dfrac {T_{n+1}}{T_n} } = \lim_{n\rightarrow \infty} {\dfrac {T_n+T_{n-1}+T_{n-2}} {T_n} } \\ \quad = \displaystyle \lim_{n\rightarrow \infty} {\left( 1 + \dfrac {T_{n-1}} {T_n} + \dfrac {T_{n-2}T_{n-1}} {T_{n-1} T_n} \right)} = 1+\dfrac{1}{y} + \dfrac{1}{y^2} \\ \Rightarrow y^3-y^2-y-1=0 \quad ...(1)$

Solve the cubic equation Eq.1 by Cardano's Formula as follows:

Substituting $\space x=t+\frac {1}{3} \space \Rightarrow$ Eq.1: $\quad t^3 - \frac {4}{3} t - \frac {38}{27} = 0\quad...(2)$

Let $\space t = u + v \space \Rightarrow$ Eq.2: $\space u^3+v^3+(3uv-\frac{4}{3})(u+v)-\frac {38}{27} = 0\quad...(3)$

Now let $\space 3uv-\frac{4}{3} = 0\space \Rightarrow$ Eq. 3: $\space u^3+v^3 = \frac {38}{27} \space$ also $\space u^3v^3 = \frac {64}{729}$

Therefore, $\space u^3 \space$ and $\space v^3 \space$ are roots of $\space z^2 - \frac {38}{27}z + \frac {64}{729} = 0 \quad ...(4)$

$\Rightarrow u^3, v^3 = \frac {19}{27} \pm \sqrt{\frac {19^2}{27^2} - \frac{64}{729}} = \frac {19}{27} \pm \frac {\sqrt{33}}{9} = \frac {1}{27} (19 \pm 3\sqrt{33} ) \\ \Rightarrow x = t + \frac {1}{3} = \frac {1}{3} + u + v = \frac {1}{3} ( 1 + \sqrt [3] {19 - 3\sqrt{33}} + \sqrt [3] {19 + 3\sqrt{33}} )$

$\Rightarrow A+B+C+D+E+F+G+H+I+J \\ \quad = 1+5+2+3+1+19+3+33+19+3+33 = \boxed {122}$