<s>[INST] Solution as a quadratic equation If I know that $y$ is a solution to $f(x) = x^3 + 3x + 1$ in the complex numbers, I want to find $1/(1+y^2)$. I have expressed this as $-y/(2y+1)$, but I think this can be simplified further to obtain a quadratic equation. How would I do this?
 [/INST] Let $${1\over1+y^2}=ay^2+by+c$$ Then $$1=(y^2+1)(ay^2+by+c)=ay^4+by^3+(a+c)y^2+by+c$$ Now $y^3=-3y-1$, $y^4=-3y^2-y$, so $$a(-3y^2-y)+b(-3y-1)+(a+c)y^2+by+c-1=0$$ which is $$(-2a+c)y^2+(-a-2b)y-b+c-1=0$$ so we get the system $-2a+c=0$, $-a-2b=0$, $-b+c-1=0$. So $a=-2b$, $c=b+1$, $4b+b+1=0$, $b=-1/5$, $a=2/5$, $c=4/5$, and we get $${1\over1+y^2}={1\over5}(2y^2-y+4)$$
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