minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$ 
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of  $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$

what i try 
If $x\neq 0,$ Then  divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\displaystyle x+\frac{1}{x}=t,$ then $\displaystyle x^3+\frac{1}{x^3}=t^3-3t$
$$f(t)=\frac{(t-1)^3}{t^3-3t}$$
How do i solve it without derivatives
Help me please 

You have some mistakes in $f(t)$. Correct is, it should be:
$$f(t) = \frac{(t-1)^3}{t^3-3t\color{red}{-1}}$$
and it's important to notice that $|t| \geq 2$ for any real $x$. So, we have to minimize $f:(-\infty,-2] \cup [2,\infty) \to \mathbb{R}$ with
$$f(t) = \frac{(t-1)^3}{t^3-3t-1}$$
We have:
$$f'(t) = \frac{3 (t-1)^2 (t^2-2t-2)}{(t^3-3t-1)^2}$$
$f'$ has only one real root in the definition domain of $f$ and that is $t_0=1+\sqrt{3}$. Also $f'(t) < 0$ for $t < t_0$ and $f'(t) > 0$ for $t > t_0$, so $t$ is a local minima. Therefore:
$$\min f(t)=f(1+\sqrt{3})=2\sqrt{3}-3$$
This is attained when $x+\frac{1}{x} = 1+\sqrt{3}$, which gives two values for $x = \frac{1}{2}(1 \pm \sqrt[4]{12} + \sqrt{3})$.
