Newton‑Sum / Symmetric‑Sum Technique

One‑Minute Solution

For a monic cubic, Vieta’s formulas state

$$ \begin{aligned} r_1+r_2+r_3 &= -a,\ r_1r_2+r_2r_3+r_3r_1 &= b,\ r_1r_2r_3 &= -c. \end{aligned} $$

We already know the sum of roots $S_1=r_1+r_2+r_3=3$.

The sum of squares is linked to the pairwise‐product sum by the identity

Plug in the given numbers:

$$ 5 = 3^{2} - 2\bigl(r_1r_2+r_2r_3+r_3r_1\bigr) ;\Longrightarrow; 5 = 9 - 2b ;\Longrightarrow; 2b = 4 ;\Longrightarrow; b = 2. $$

(No need to solve for the individual roots; the symmetric‑sum identity did all the work.)