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?
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
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