Is this even solvable?

By Vieta's formula, we know that $a+b+c=0$ . Therefore,

$\left|\frac{2}{251} \left((a+b)^3 + (b+c)^3 + (c+a)^3 \right) \right|$ $= \left|\frac{2}{251} \left(-a^3 -b^3 -c^3 \right) \right|$ $= \left|\frac{2}{251} \left((75a-251) + (75b-251) + (75c-251) \right) \right|$ $= \left|\frac{2}{251} \left(75(a+b+c) - 3*251 \right) \right|$ $= \left|\frac{2}{251} \left(-3*251 \right) \right|$ $= \left|-6 \right|$ $= \boxed{6}$

As, $a+b+c=0, abc=251$ , then $a^3+b^3+c^3=3abc$ .

Now, $|\frac{2}{251}(a+b)^3+(b+c)^3+(c+a)^3|$

$= \frac{2}{251}|(-c)^3+(-a)^3+(-b)^3|$

$= \frac{2}{251}|a^3+b^3+c^3|$

$= \frac{2}{251}3abc$

$=6$

By Vieta's formulas, we get the following two equations: $a+b+c=0, ~~abc=251.$ Consequences: $b+c=-a, ~c+a=-b, ~a+b=-c$ . Now: $\begin{aligned} \sum_{\text{cyc}} (a+b)^3 &=2(a^3+b^3+c^3)+3ab(a+b)+3bc(b+c)+3ca(c+a) \\ &=2(a^3+b^3+c^3)-9abc \end{aligned}$ Also from the trivial result $a+b+c=0 ~\implies a^3+b^3+c^3=3abc$ , we have: $\sum_{\text{cyc}} (a+b)^3=2(a^3+b^3+c^3)-9abc=-3abc=-3\times 251$ Hence our desired answer is: $\left|\dfrac{2}{251}\cdot\left(\sum_{\text{cyc}} (a+b)^3\right)\right|=\left|\dfrac{2}{251}\times (-3)\times 251\right|=\left|-6\right|=\boxed{6}$

a = 2.990187592 0

b = -1.495093796 -9.039132501

c = -1.495093796 9.039132501

Checked that only 1 result for several cyclic orders,

-363.132034703923+677.934937601324i

-26.7359305921814+9.82664521666655E-15i

-363.132034703923-677.934937601324i

-753.000000000027

|-6| = 6