Solve cubic polynomial

Question

How can I solve this third-degree polynomial? I want to solve it for $y$.

$x=y^3+y-9$

I can simplify it to $x+9=y(y^2+1)$ but I don't get any further.

Answer 1

That's a job for Cardano's formula. First, however, we have to ensure the function is invertible: if $f(y)=y^3+y-9$, then $f'(y)=3y^2+1$ and, as this is everywhere positive, we're done.

Next rewrite to $y^3+y-9-x=0$, to get $$ y=\sqrt[3]{\frac{x+9}{2}+\sqrt{\frac{1}{27}+\frac{(x+9)^2}{4}}}+ \sqrt[3]{\frac{x+9}{2}-\sqrt{\frac{1}{27}+\frac{(x+9)^2}{4}}} $$