# Value of Plastic Constant

## Theorem

The plastic constant $P$ is evaluated as:

 $\ds P$ $=$ $\ds \sqrt [3] {\frac {9 + \sqrt {69} } {18} } + \sqrt [3] {\frac {9 - \sqrt {69} } {18} }$ $\ds$ $=$ $\ds 1 \cdotp 32471 \, 79572 \, 44746 \, 02596 \, 09088 \, 54 \ldots$

## Proof

By definition, the plastic constant $P$ is the real root of the cubic:

$x^3 - x - 1 = 0$

Recall Cardano's Formula:

Let $P$ be the cubic equation:

$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$

Then $P$ has solutions:

$x_1 = S + T - \dfrac b {3 a}$
$x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
$x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

where:

$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$

where:

$Q = \dfrac {3 a c - b^2} {9 a^2}$
$R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$

Here we have:

 $\ds a$ $=$ $\ds 1$ $\ds b$ $=$ $\ds 0$ $\ds c$ $=$ $\ds -1$ $\ds d$ $=$ $\ds -1$

Hence:

 $\ds Q$ $=$ $\ds \dfrac {3 \times 1 \times \paren {-1} - 0^2} {9 \times 1^2}$ $\ds$ $=$ $\ds \dfrac {-3} 9$ $\ds$ $=$ $\ds -\dfrac 1 3$ $\ds R$ $=$ $\ds \dfrac {9 \times 1 \times 0 \times \paren {-1} - 27 \times 1^2 \times \paren {-1} - 2 \times 0^3} {54 \times 1^3}$ $\ds$ $=$ $\ds \dfrac {27} {54}$ $\ds$ $=$ $\ds \dfrac 1 2$

and so:

 $\ds \sqrt {Q^3 + R^2}$ $=$ $\ds \sqrt {\paren {-\dfrac 1 3}^3 + \paren {\dfrac 1 2}^2}$ $\ds$ $=$ $\ds \sqrt {\dfrac 1 4 - \dfrac 1 {27} }$ $\ds$ $=$ $\ds \sqrt {\dfrac {27 - 4} {108} }$ $\ds$ $=$ $\ds \sqrt {\dfrac {3 \times 23} {3 \times 2^2 \times 3^3} }$ $\ds$ $=$ $\ds \sqrt {\dfrac {69} {18^2} }$ $\ds$ $=$ $\ds \dfrac {\sqrt {69} } {18}$ $\ds \leadsto \ \$ $\ds S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$ $=$ $\ds \sqrt [3] {\dfrac 1 2 + \dfrac {\sqrt {69} } {18} }$ $\ds$ $=$ $\ds \sqrt [3] {\dfrac {9 + \sqrt {69} } {18} }$ $\ds T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$ $=$ $\ds \sqrt [3] {\dfrac 1 2 - \dfrac {\sqrt {69} } {18} }$ $\ds$ $=$ $\ds \sqrt [3] {\dfrac {9 - \sqrt {69} } {18} }$

Then:

 $\ds S + T - \dfrac b {3 a}$ $=$ $\ds \sqrt [3] {\dfrac {9 + \sqrt {69} } {18} } + \sqrt [3] {\dfrac {9 - \sqrt {69} } {18} } - \dfrac 0 {3 \times 1}$ $\ds$ $=$ $\ds \sqrt [3] {\dfrac {9 + \sqrt {69} } {18} } + \sqrt [3] {\dfrac {9 - \sqrt {69} } {18} }$

The number can then be calculated.

Since $S \ne T$, the other two roots $x_2, x_3$ has non-zero imaginary parts $\pm \dfrac {i \sqrt 3} 2 \paren {S - T}$.

Hence the root above is the only real root.

$\blacksquare$