Please help, equation $3x-x^3=1$ has three roots. Interesting fact that $|x_3|= x_1+x_2$. Is it possible to reduce this equation to a quadratic?
Reduce specific cubic equation to quadratic
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What we can do is use the root coefficient relationships to express any two roots of the cubic equation in terms of the third root. Let the cubic equation be $ax^3+bx^2+cx+d=0$ and let $r$ be any root. Then the remaining two roots must sum to $-(ra+b)/a$ and their product must be $-d/(ra)$ leading to the quadratic equation:
$(ra)x^2+(r^2a+rb)x-d=0$
and the other two roots are:
$r_{\pm} = \frac{-(r^2a+rb)\pm \sqrt{(r^2a+rb)^2+4rad}}{2ra}$
In some applications, such as the thermodynamics problem of modeling vapor liquid equilibrium with a cubic equation of state, we can set up cubic equations to have a predefined root (corresponding to one phase in the equilibrium problem) and use the above scheme to get the other roots.
No, because our equation has three roots.
If you mean if is it possible to write
this equation in the form $(x-a)(x^2+bx+c)=0$, where $\{a,b,c\}\subset\mathbb Q$,
then it's impossible again because our equation has no rational roots.