Let $ax^3+bx^2+cx+d$ be a cubic polynomial. Let $r_1$, $r_2$, $r_3$ be its roots. Let us assume all three roots are real.
It is well known that $r_1+r_2+r_3=-\dfrac{b}{a}$.
Is there any way to calculate $|r_1|+|r_2|+|r_3|$ without actually computing the roots?
The answer is of course easy if all roots positive or all roots are negative. What about otherwise?
Thank you.
Probably there is no simple formula that covers all cases without explicitly computing at least one root. Fortunately, we are able to compute roots of a cubic equation via algebraic techniques. A variety of methods is given by Wikipedia. Note that if there are only real roots, they can be rendered only in terms of trigonometric functions (unless you want to deal with complex variables), so pay close attention to those.
If you know that the cubic equation $ax^3+bx^2+cx+d$ has only real roots, then the trigonometric formula that applies in this case is equivalent to a formula involving the local minimum and maximum values of the cubic polynomial. Render
$M_1=P((-b+\sqrt{b^2-3ac})/3a)$
$M_2=P((-b-\sqrt{b^2-3ac})/3a)$
where $P(x)=ax^3+bx^2+cx+d$. Then the roots are given by
$r_1,r_2,r_3=-\dfrac{b}{3a}+\dfrac{2\sqrt{b^2-3ac}}{3a}\sin(\frac{1}{3}(\arcsin\dfrac{M_2+M_1}{M_2-M_1}+2k\pi)),k\in\{0,\pm1\}.$
Note that the arcsine will be within its domain when the $M$ values are opposite in sign (or when one of them is zero in the degenerate case). This corresponds to the local minimum and maximum points straddling the zero line so that these are then sure to be sandwiched between three real roots.