Question: **$\{.\}$ represents fractional part function**

$$f’(x)=3x^2-3=0$$
$$\implies x=\pm 1$$

Function is maximum at $x=-1$ and minimum at $x=1$

$$f(-1)=2$$ and $$f(1) =-1$$

One root lies between $-1$ and $1$ and the others lie before and after the numbers respectively.

$$\{x\_1\}+\{x\_2\}+\{x\_3\}$$
$$=(x\_1+x\_2+x\_3)-([x\_1]+[x\_2]+[x\_3])$$
$$3-([x\_1]+0+[x\_3])$$

I don’t know how far $x\_1$ and $x\_3$ are from $-1$ and $1$, so the value of their greatest integer can’t be determined. How do I proceed?

Answer: 
According to the intermediate value theorem,If you care about $x=-2,-1,0,1,2$ you can find the answer
$$f(-2)\times f(-1)<0 \to \exists x\_3|x\_3 \in (-2,-1), f(x\_3)=0\\ \to -2<x\_3<-1 \to [x\_3]=-2$$ also
$$f(0)f(1)<0 \to \exists x\_2|x\_2 \in (0,1), f(x\_2)=0\\ \to 0<x\_2<1 \to [x\_2]=0 $$ also
$$f(1)f(2)<0 \to \exists x\_1|x\_1 \in (1,2), f(x\_1)=0\\ \to 1<x\_3< \to [x\_3]=1$$ so
$$x^3+0x^2-3x+1=(x-x\_1)(x-x\_2)(x-x\_3)=x^3-\underbrace{(x\_1+x\_2+x\_3)}\_{\to 0}x^2+...$$ note that sum of roots is zero  

so $$=(x\_1+x\_2+x\_3)-([x\_1]+[x\_2]+[x\_3])=0-(-2+0+1)=1$$
Easy to see that: $x\_1\in(-2,-1),$ $x\_2\in(0,1)$ and $x\_3\in(1,2).$

Thus, since $\{x\}=x-[x],$ we obtain:
$$\{x\_1\}+\{x\_2\}+\{x\_3\}=2+x\_1+x\_2+x\_3-1=1.$$