wrapping zero

x^3-2x+1=x(x^2-x-1)+x^2-x+1 {solving the equation we get x^3-2x+1}
=x*0 +x^2-x+1=x^2-x+1 x^2-x+1=1(x^2-x-1)+2
=0+2=2

$x^2-x-1=0$ , so $x^2-x=1$ and $x^2=x+1$ , and $x^3=x^2+x$ .

$\Rightarrow x^3-2x+1=x^2-x+1=x+1-x+1=\boxed{2}$ .

Of course we have

$x \cdot 0 = -0$

Since $0 = x^2 - x -1$ we have

$x \left( x^2 - x -1\right) = -\left(x^2 - x -1\right) \Longrightarrow x^3 -x^2 -x = -x^2 + x +1$

hence

$x^3 - 2x = 1 \Longrightarrow x^3 - 2x +1 = 2.$

$x^2 - x -1 = 0 \Rightarrow (x+1)(x^2 -x-1) = 0(x+1)$

So, $x^3 -2x -1 = 0 \Rightarrow x^3 - 2x = 1$

Therefore, $x^3 -2x + 1 = (x^3 -2x) + 1 = 1+1 = 2$

$x^2-x-1=0$ $››$ $x^2=x+1$ $x^3=x(x+1)$

$x^3-2x+1$ $=x(x+1)-2x+1$ $=x^2-x+1$ $(As, x^2-x-1=0 , so, x^2-x=1)$ $››››››$ $1+1$ = $2$

x^{3}-2x+1=x^{2}-x-1+x(x^{2}-x-1)+2=0+x(0)+2=0+0+2=2