$x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}$. Then what is the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\}$?

Question

If the equation $x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}\;,$ Where $x_{1}<x_{2}<x_{3}$.

Then the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\} = \;,$ Where $\{x\}$ Represent fractional part of $x.$

$\bf{My\; Try::}$ Let $f(x) = x^3-3x+1\;,$ Then Using Second Derivative Test, We get

$f'(x) = 3x^2-3$ and $f''(x) = 6x\;,$ Now for $\max$ and $\min\;,$ Put $f'(x) =0\;,$ So we get

$3x^2-3=0\Rightarrow x = \pm 1.$ Now $f''(1) = 6>0$

So $x=1$ is a Point of $\min$ and $f''(-1) = -6<0$

So $x=-1$ is a Point of $\max.$

And Rough graph of $f(x) = x^3-3x+1$ is Like

So we get $-2<x_{1}<-1$ and $0<x_{2}<1$ and $1<x_{3}<2$

So $\lfloor x_{1}\rfloor = -2$ and $\lfloor x_{2}\rfloor = 0$ and $\lfloor x_{3}\rfloor = 1$

Now I did not Understand How can I calculate value of fractional part

Help me, Thanks

Answer 1

Note that $$\{x\}=x-\lfloor x\rfloor$$ and that $$x_1+x_2+x_3=0.$$

Answer 2

$\left\{x_1\right\}+\left\{x_2\right\}+\left\{x_3\right\}=x_1+x_2+x_3-[x_1]-[x_2]-[x_3]=0-(-2+0+1)=1$