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 
 A: Hint:
$a=x^2$
$b=-x$
$c=1$ $$Min{\frac{(a+b+c)^3}{a^3+b^3+c^3}}=?$$
A: 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 $x = \frac{1}{2}(1 \pm \sqrt[4]{12} + \sqrt{3})$.
A: 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.
A: A solution without derivatives
Clearly, $t \ge 2$ or $t \le -2$. 
Denote the minimum of $f(t)=\frac{(t-1)^3}{t^3-3t-1}$ on $|t|\ge 2$ by $m > 0$.
Since $f(3) = \frac{8}{17}$, we have $m \le \frac{8}{17}$.
Note that $f(t) - 1 = \frac{3t(2-t)}{t^3-3t-1} \ge 0$ for $t \le -2$.
Thus, the minimum occurs on $t \ge 2$.
$f(t) = m$ leads to a cubic equation $g(t) = 0$ where
\begin{align}
g(t) &= (t^3-3t-1)(f(t)-m)\\
&= (1-m)t^3 - 3t^2 + (3+3m)t + m - 1.
\end{align}
Clearly, $g(t) = 0$ has a real root on $t \ge 2$.
Since $g(0) = m - 1 < 0$ and $g(2) = 1 - m > 0$, we know that $g(t) = 0$ has a real root on $(0, 2)$.
Thus, we can factor $g(t) = (1-m)(t-t_1)(t-t_2)(t-t_3)$ for some $t_1 \in (0, 2)$ and $t_2 \ge 2$.
Since $g(t) \ge 0$ for $t \ge 2$, it is easy to prove that $t_2 = t_3$.
Thus, we have $g(t) = (1-m)(t-t_1)(t-t_2)^2$. 
Since $g(t) = 0$ has a multiple root, its discriminant is zero, 
i.e. $\Delta = 81(m^2+6m-3)m^2 = 0$ which results in $m = 2\sqrt{3} - 3$.
Finally, let us prove that $m = 2\sqrt{3} - 3$ is indeed the minimum of $f(t)$ on $|t|\ge 2$.
It is easy to check that
$$f(t) - (2\sqrt{3} - 3) = \frac{1}{t^3 - 3t - 1}(4-2\sqrt{3})\Big(t - 1 + \tfrac{\sqrt{3}}{2}\Big)\Big(t - 1 - \sqrt{3}\Big)^2$$
and hence $f(t) - (2\sqrt{3} - 3) \ge 0$ for $|t| \ge 2$, with equality if and only if $t = 1 + \sqrt{3}$.
We are done.
