<s>[INST] How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.
 [/INST] $$x^5+x^3+x-3=0\\(x^5-1)+(x^3-1)+(x-1)=0\\(x-1)(x^4+x^3+2 x^2+2x+3)=0$$
So the solution you wanted is $x=1$.
Sorry but it is just a solution ad hoc.
</s>