x^3 = 2x+1
$$x^3 -2x-1 = 0 \qquad \Rightarrow x_1=-1 \\ \\
(x^3 -2x-1) : (x+1) = x^2-x-1 \\\\
(x+1)\underbrace{(x^2-x-1)}_{=0} = 0\\\\
x^2-x-1 = 0\\
x_{2,3}=\frac{
1\pm\sqrt{1-1*4*(-1)}
}
{2*1} = \frac{
1\pm\sqrt{5 }
}
{2} \\ \\
x_2 = \frac{1+\sqrt{5}}{2} =1.61803398875 \\\\
x_3 = \frac{1-\sqrt{5}}{2} =-0.61803398875$$
x^3 = 2x+1 subtract everything on the right side to make it 0....so we have....
x^3 - 2x - 1 = 0 and from the Facror Theorem, we have that -1 is a root (solution)
Therefore, using a little synthetic division, we have
-1 [ 1 0 - 2 -1 ]
- 1 1 1
-----------------------
1 -1 -1 0
This tells us that the polynomial remaining after we divide x^3 - 2x -1 by (x + 1) = x^2 - x - 1
And setting this = 0, we have that x =
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$
So these are the 3 solutions......BTW.....the last 2 solutions are better known as "-phi" and "Phi"......
x^3 = 2x+1
$$x^3 -2x-1 = 0 \qquad \Rightarrow x_1=-1 \\ \\
(x^3 -2x-1) : (x+1) = x^2-x-1 \\\\
(x+1)\underbrace{(x^2-x-1)}_{=0} = 0\\\\
x^2-x-1 = 0\\
x_{2,3}=\frac{
1\pm\sqrt{1-1*4*(-1)}
}
{2*1} = \frac{
1\pm\sqrt{5 }
}
{2} \\ \\
x_2 = \frac{1+\sqrt{5}}{2} =1.61803398875 \\\\
x_3 = \frac{1-\sqrt{5}}{2} =-0.61803398875$$