Can you manipulate this?

It is given that $x^2 - x - 1 = 0$ $\implies \color{#3D99F6}{x^2 = x+1}$ $\implies \color{#D61F06}{x^3} = \color{#3D99F6}{x^2} + x = \color{#3D99F6}{(x+1)} + x \color{#D61F06}{= 2x+1}$ .

Therefore, $\color{#D61F06}{x^3}-2x+1 = \color{#D61F06}{(2x+1)} -2x +1 = \boxed{2}$ .

Let $x^3 - 2x+1=S$
Then; $x^3+1=S+2x$
$\implies (x+1)(x^2-x+1)=S+2x$
(Since, $\color{#3D99F6}{a^3 + b^3 = (a+b)(a^2-ab+b^2)}$ )
$\implies2(x+1) = S +2x$
(Since, $\color{#3D99F6}{x^2-x-1=0; ==> x^2-x+1=2}$ )
$\implies 2x + 2 = S+2x$
Hence, $x^3-2x+1=S=\boxed{2}$

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

$= (0)(x+1)+2$

$=2$

The given expression can be be rearranged as

$x^3-2x-1 = x(x^2-2)+1= x(x+1-2)+1$ as $x^2-x-1 = 0 \Rightarrow x^2 = x+1$

$= x^2-x +1 = x^2-x-1+2 = 2$ as $x^2-x-1= 0$

Hence the answer is 2.

$\\ x^3 - 2x + 1 = \\\\ (x^3 + 1) - 2x = \\\\ \left [ (x + 1)(x^2 - x - 1) + 2x + 2 \right ] - 2x = \\\\ \left [ (x + 1) \cdot 0 + 2x + 2 \right ] - 2x = \\\\ 0 + 2x + 2 - 2x = \\\\ \boxed{2}$

$x²-x-1=0$ $..........(1)$

$x²=x+1$

Multiply both side with $x$

$x³=x²+x$ $..........(2)$

Otherwise, from $(1)$ we get

$x = x²-1$

Adding $x$ on both sides we get

$2x = x² + x -1$ $..........(3)$

Now,

$x³ -2x+1 = (2) - (3) +1$

$= x² + x - (x² + x -1) +1$

$= x² +x - x² -x +1 +1$

$= \boxed{2}$