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 

Hint:
$a=x^2$
$b=-x$
$c=1$ $$Min{\frac{(a+b+c)^3}{a^3+b^3+c^3}}=?$$
