I was trying to solve this polynomial
$$x(3-x^2)=1$$
I worked for the term $(3-x^2)$, I thought that this term cannot be $0$, thus
$$3-x^2 >0$$
$x< \sqrt{3}$, $x<-\sqrt{3}$ is rejected because then it will be multiplied by -ve $x$ and having -ve result.
I really want to know what is the approach to solve such question.
Consider the polynomial $$x^3+ax^2+bx+c=0$$ By putting $x=y-\frac{a}3$ $\,\,$ we get:$$y^3+(b-\frac{a^2}{3})y+\frac{2a^3}{27}-\frac{ab}{3}+c=0$$ Let:$p=b-\frac{a^2}{3}$ and $q=\frac{2a^3}{27}-\frac{ab}{3}+c$, then we have: $$y^3+py+q=0$$ Now let: $y=r+s$, thus: $$y^3=r^3+s^3+3rs(r+s)=r^3+s^3+3rsy$$ So we should have: $$y^3-3rsy-(r^3+q^3)\equiv y^3+py+q$$ It's mean:
$$
rs=\frac{-p}{3}\,\,\,,\,\,\,
r^3+s^3=-q
$$
Which can be written as following: $$
r^3s^3=\frac{-p^3}{27}\,\,\,,\,\,\,
r^3+s^3=-q
$$ So, $r^3 \,, s^3$ are the roots of following equation: $$t^2+qt-\frac{p^3}{27}=0$$ i.e: $$r^3=-\frac{q}2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}$$
$$s^3=-\frac{q}2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}$$ And finally:
$$
\begin{align}
x=y-\frac{a}3&=r+s-\frac{a}3\\
&=\sqrt[3]{-\frac{q}2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}-\frac{a}{3}
\end{align}
$$
