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.
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1Are you trying to invert $f(y)=y^3+y-9$? – Shubham Johri Dec 15 '18 at 11:59
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Need cubic equation. [like quadratic but more involved]. Maybe try "Cardano's formula", – coffeemath Dec 15 '18 at 11:59
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You made a mistake. It is $x+9=y(y^2+1)$. Why do you write $x$? Is it another variable? – Dietrich Burde Dec 15 '18 at 12:01
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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 $$ y=\sqrt[3]{\frac{x+9}{2}+\sqrt{\frac{1}{27}+\frac{(x+9)^2}{4}}}+ \sqrt[3]{\frac{x+9}{2}-\sqrt{\frac{1}{27}+\frac{(x+9)^2}{4}}} $$
egreg
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And there is only one solution as the derivative is $3y^2+1$ ie positive.... – dmtri Dec 15 '18 at 12:13
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@dmtri Yes: the discriminant (what's under the square root) is positive, so there's no choice in order to obtain the unique real root. – egreg Dec 15 '18 at 12:30