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 

Here is a 'non-calculus' solution.
We want to minimize the function $f:(-\infty,-2] \cup [2,\infty) \to \mathbb{R}$
$$f(t) = \frac{(t-1)^3}{t^3-3t-1}$$
Let's pretend we don't know calculus and that we're really clever to spot that for $t=1+\sqrt{3}$, we have $f(t)=2\sqrt{3}-3$. So, we will prove:
$$\frac{(t-1)^3}{t^3-3t-1}\geq 2\sqrt{3}-3$$
If $t \leq -2$, we have $6t-3t^2<0$ and $t^3-3t-1 < 0$, so:
$$\frac{(t-1)^3}{t^3-3t-1} = 1+\frac{6t-3t^2}{t^3-3t-1} > 1 > 2\sqrt{3}-3$$
If $t \geq 2$, we have $t^3-3t-1 > 0$, so it will be enough to prove:
$$(t-1)^3 \geq (2\sqrt{3}-3)(t^3-3t-1)$$
Since we know $t=1+\sqrt{3}$ is the equality case, this is easy to factor into:
$$(t-1-\sqrt{3})^2\left[(4-2\sqrt{3})t+4\sqrt{3}-7\right]\geq 0$$
and this is obviously true, because
$$(4-2\sqrt{3})t+4\sqrt{3}-7 \geq 2(4-2\sqrt{3})+4\sqrt{3}-7=1 > 0$$
completing the proof.
