# 塑膠數

 ${\displaystyle {\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}+{\sqrt[{3}]{{\frac {1}{2}}-{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}}$ 二進位 約為1.010100110010000010110111010011101100101 八進位 約為1.2462026723545104533260274211370405060463 十進位 約為1.324717957244746025960908854478097340734 十六進位 約為1.5320B74ECA44ADAC178897C41461334737F8172F

${\displaystyle {\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}+{\sqrt[{3}]{{\frac {1}{2}}-{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}}$

## 塑膠數的來源

${\displaystyle x^{3}-x-1=0\,}$

${\displaystyle x={\frac {\lambda }{y}}+y\,}$

${\displaystyle y={\frac {1}{2}}{\sqrt {x^{2}-4\lambda }}\,}$

${\displaystyle -1-y-{\frac {\lambda }{y}}+\left(y+{\frac {\lambda }{y}}\right)^{3}=0\,}$

${\displaystyle y^{6}+y^{4}\left(3\lambda -1\right)-y^{3}+y^{2}\left(3\lambda ^{2}-\lambda \right)+\lambda ^{3}=0\,}$

${\displaystyle \lambda ={\frac {1}{3}}\,}$，將其帶入上面方程，并設${\displaystyle z=y^{3}\,}$，得到一個${\displaystyle z}$二次方程

${\displaystyle z^{2}-z+{\frac {1}{27}}=0\,}$

${\displaystyle z={\frac {1}{18}}\left(9+{\sqrt {69}}\right)\,}$

${\displaystyle y^{3}={\frac {1}{18}}\left(9+{\sqrt {69}}\right)\,}$

${\displaystyle y}$有實數解

${\displaystyle y={\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}\,}$

${\displaystyle x={\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}+{\sqrt[{3}]{{\frac {1}{2}}-{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}\,}$