Tribonacci Constant is Unique Real Root of x^3-x^2-x-1

Definition
The Tribonacci constant $\eta$ is the one real root of the cubic:
 * $x^3 - x^2 - x - 1 = 0$

Its decimal expansion starts:
 * $\eta = 1 \cdotp 83928 \, 67552 \, 1416 \ldots$

Initial substitution
We make a sign rectification because this cascade works with only one substitution at a time.

We choose $+$ because we know $x^3 - x^2 - x$ has a zero root and a positive root, so $x^3 - x^2 - x - 1$ can only have a positive root.

The cascade will account for the $-$ solution anyway.

Lambda substitution
This next part of the substitution occurs because of the following logic:

$P1:$ To take a cube root of a real number, there must be three possible solutions (one real, two complex).

$P2:$ This polynomial is being restricted to the real xy-plane, i.e. no complex roots are allowed.

$C1:$ To take the cube root of this real number, it needs to vary by signs as the two complex would.

$P3:$ A complex number has its own arguments with which sign works, i.e. $+i$ vs. $-i$, which does not apply on the real number line.

$C2:$ All combinations of signs have to be represented, and there must be three.

$P4:$ Since a sign has two states, $+$ and $-$, there must be at least one other entity able to be modified with a sign.

$C3:$ To take the cube root of this real number, two parts of it must be able to be modified with a sign together exactly thrice.

We can see that the first cube root is trivially determined:

$v^3 = \dfrac \Lambda {27}$

$v = \dfrac 1 3 { \Lambda }^{ \dfrac 1 3 }$

The two parts we can modify are $\dfrac 1 3$ and $\Lambda$. We already have $+$ for the first and $+$ for the second.

We know $-1 × -1 = +1$, so we can have $-$ for the first and $-$ for the second:

$v = - \dfrac 1 3 \paren { - \Lambda }^{ \dfrac 1 3 }$

There must be a third way. We can have both negative signs on the first part or on the second part (as a cube root). It's possible to write it as both at the same time if it is in the form of a third part, so that it can multiply by either:

$v = \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } { \Lambda }^{ \dfrac 1 3 }$

Any of these three roots are able to be used in the cascade.

Reverse substitution
Substitute for $u = v + \dfrac w v$ in each of the three roots, where $w = \dfrac 4 9$.

$u = \dfrac 4 3 \paren { \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { \Lambda }^{ \dfrac 1 3 }$

$u = - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren { \Lambda }^{ \dfrac 1 3 }$

$u = \dfrac 4 3 \paren { -1 }^{ \dfrac 2 3 } \paren { \Lambda }^{ - \dfrac 1 3 } - \dfrac 1 3 \paren { - \Lambda }^{ \dfrac 1 3 }$

We can pick any of these three roots. The following will use the second of these three.

For reference, let ${ \overline \Lambda } = 19 - 3 \sqrt {33}$, the negative form.

Part 4: Lambda cubic cascade
Therefore, the Tribonacci constant is equal to the preceding. It can be evaluated manually or with a calculator to be approx. $1.83929$.