How to find interval where function $f(x)=x+\frac{1}{x^{3}}$ is one to one\injective? How to find interval where function is one to one $f(x)=x+\frac{1}{x^{3}}$ ?(graphically or algebraically analytically)
 Let $f(x) =f(y)$ this gives $ (x-y)(\frac{(xy)^{3}-y^{2}-x^{2}+xy}{(xy)^{3}})=0 $
I don't know how to proceed further.

Hint :
$$f'(x) = 1 - \frac{3}{x^4} \implies \begin{cases} f'(x) > 0, \; x \in (-\infty,-3^{1/4})\cup(3^{1/4},\infty) \\ f'(x) <0, \; x \in (-3^{1/4},0)\cup(0,3^{1/4})  \end{cases}$$ 
