should I search $f^{-1}(x)$ or is there an easier way to solve it? $\forall x\in\mathbb R : f(x) = x^3 +x -8$
solve : $2f(x) +3f^{-1}(x) =10 $
I actually tried to write it as : $f^{-1}(x) = \frac{10-2f(x)}{3}$ 
Hence : $x=f(\frac{10-2f(x)}{3})$ 
But it seems to be so hard to solve , do you have any suggestions for solving this problem ?

With $y:=f^{-1}(x)$, the equation becomes
$$\tag12f(f(y)) +3y=10$$which produces an awfully high degree equation:
$$ 2y^9 + 6y^7 - 48y^6 + 6y^5 - 96y^4 + 388y^3 - 48y^2 + 389y - 1066=0.$$
Solving such an equation exactly is beyond hope, in general. By sheer luck we may find a solution by trying a few small integer values for $y$, and indeed $y=2$ turns out to be a solution. Incidentally, we find $x=f(y)=2$ as well and face-palm heavily.
We are still left with the question whether there are more solutions coming from the seemingly untractable remaining degree 8 factor. 

So let's start over again.
As $(1)$ involves $f$ applied twice, it seems useful to investigate iterates of $f$ in general, and the most prominent features of iterates: fixpoints.
We observe that $$f(t)-t=t^3-8 $$
is positive for $t>2$, negative for $t<2$, and zero (only) for $t=2$.
It follows that $f(f(y))\gtreqless f(y)\gtreqless y\gtreqless$ if $y\gtreqless 2$, thus making $$2f(f(y))+3y\gtreqless 2\cdot 2+3\cdot 2=10\qquad\text{if }y\gtreqless2. $$
We conclude that $(1)$ has exaclt yone solution $y=2$ (and then also $x=2$).
