$ f(x)= {x^3}-2x+1. $ We want to show that $ \lim_{x\to 2} ({x^3}-2x+1)= 5 $. So here, $ \lvert f(x)-f(2)\rvert = \lvert {x^3}-2x+1-5\rvert$ = $ \lvert {x^3}-2x-4\rvert $.
$ \lvert ({x^3}-2x)+(-4)\rvert \leq \lvert {x^3}-2x\rvert + \lvert -4\rvert = \lvert {x^3}-2x\rvert + 4 $, by triangle inequality. I'm confused on what to do next for the $ \epsilon$ and $\delta$.
As you have computed, we have $$|f(x)-f(2)|=|x^3-2x-4|.\tag{0}$$ Now, we have $$x^3-2x-4=(x-2)(x^2+2x+2)=(x-2)[(x+1)^2+1].\tag{1}$$ If $|x-2|<1$, then $1<x<3$ and we can bound $$0<(x+1)^2+1\leq 16+1=17.\tag{2}$$
Therefore, for any $\epsilon>0$, we can choose $\delta=\min\{ 1, \epsilon/17\}>0$ such that if $|x-2|<\delta$, then $$|f(x)-f(2)|=|x^3-2x-4|~~\mbox{ by }(0)\\ =|x-2|\cdot|(x+1)^2+1|~~\mbox{ by }(1)\\ \leq \frac{\epsilon}{17}\cdot 17 ~~\mbox{ by }(2)\\ =\epsilon. $$