[THOUGHT] Let me start by finding the values of a, b, c from the given system of equations. Then I'll compute the integral using the residue theorem.

First, let me find a, b, c. We have:

• a + b + c = 151
• a² + b² + c² = 10939
• a³ + b³ + c³ = 957871

I can use Newton's identities to find the elementary symmetric polynomials, then find the roots. [/THOUGHT] [PYTHON] import numpy as np from itertools import combinations_with_replacement

Given conditions

s1 = 151 # a + b + c s2 = 10939 # a^2 + b^2 + c^2 s3 = 957871 # a^3 + b^3 + c^3

Elementary symmetric polynomials

e1 = a + b + c = s1

e2 = ab + bc + ca

e3 = abc

We know: s2 = e1^2 - 2*e2

So: e2 = (e1^2 - s2) / 2

e1 = s1 e2 = (e1**2 - s2) // 2 print(f"e1 = {e1}, e2 = {e2}")

We know: s3 = e1^3 - 3e1e2 + 3*e3

So: e3 = (s3 - e1^3 + 3e1e2) / 3

e3 = (s3 - e1**3 + 3e1e2) // 3 print(f"e3 = {e3}")

Now a, b, c are roots of x^3 - e1x^2 + e2x - e3 = 0

Let's find the roots

coeffs = [1, -e1, e2, -e3] roots = np.roots(coeffs) print(f"Roots: {roots}")

Round to nearest integers

a, b, c = sorted([int(round(r.real)) for r in roots]) print(f"a, b, c = {a}, {b}, {c}")

Verify

print(f"a + b + c = {a + b + c}") print(f"a^2 + b^2 + c^2 = {a2 + b2 + c2}") print(f"a^3 + b^3 + c^3 = {a3 + b3 + c3}") [/PYTHON]

[THOUGHT] Now I have a = 21, b = 33, c = 97. I need to compute the integral:

$I = \int_{-\infty}^{\infty} \frac{Q(x)e^{ix}}{(x - \frac{205}{69}i)^4} dx$

where $Q(x) = (x-21)(x-33)(x-97)$.

The pole is at $z_0 = \frac{205}{69}i$, which is in the upper half-plane (since 205/69 > 0).

For the integral $\int_{-\infty}^{\infty} \frac{f(x)e^{ix}}{(x-z_0)^4} dx$ where $z_0$ is in the upper half-plane, I can use the residue theorem by closing the contour in the upper half-plane.

The integral equals $2\pi i \cdot \text{Res}_{z=z_0} \frac{Q(z)e^{iz}}{(z-z_0)^4}$.

For a pole of order 4, the residue is: $\text{Res} = \frac{1}{3!} \lim_{z \to z_0} \frac{d^3}{dz^3}[Q(z)e^{iz}]$

Let me compute this using Python. [/THOUGHT] [PYTHON] import sympy as sp

Define variables

x = sp.Symbol('x') z = sp.Symbol('z')

Values

a, b, c = 21, 33, 97 z0 = sp.Rational(205, 69) * sp.I

Q(z) = (z-a)(z-b)(z-c)

Q = (z - a) * (z - b) * (z - c)

f(z) = Q(z) * exp(i*z)

f = Q * sp.exp(sp.I * z)

For a pole of order 4, residue = (1/3!) * d^3/dz^3 [f(z)] evaluated at z = z0

The function in the integral is f(z)/(z-z0)^4, so we need the residue of this

Residue at a pole of order n is: lim_{z->z0} (1/(n-1)!) * d^(n-1)/dz^(n-1) [(z-z0)^n * g(z)]

where g(z) = f(z)/(z-z0)^n

So for n=4: Res = (1/6) * d^3/dz^3 [f(z)] at z=z0

Compute third derivative of f

f_prime3 = sp.diff(f, z, 3) print("Third derivative of Q(z)*exp(iz):") print(f_prime3)

Evaluate at z = z0

residue = f_prime3.subs(z, z0) / 6 print("\nResidue:") print(residue)

The integral is 2pii * residue

I = 2 * sp.pi * sp.I * residue I_simplified = sp.simplify(I) print("\nIntegral I:") print(I_simplified)

Expand to get real and imaginary parts

I_expanded = sp.expand(I_simplified) I_complex = sp.N(I_expanded, 50) print("\nNumerical value of I:") print(I_complex)

Get real and imaginary parts

alpha = sp.re(I_complex) beta = sp.im(I_complex) print(f"\nalpha = {alpha}") print(f"beta = {beta}")

Compute floor(10*alpha) + floor(beta)

result = int(sp.floor(10 * alpha)) + int(sp.floor(beta)) print(f"\nfloor(10*alpha) + floor(beta) = {result}") [/PYTHON]

[THOUGHT] The computation shows that:

• α ≈ -3632.072...
• β ≈ -8.9186...

Therefore:

• floor(10α) = floor(-36320.72...) = -36321
• floor(β) = floor(-8.9186...) = -9

So floor(10α) + floor(β) = -36321 + (-9) = -36330 [/THOUGHT]

\boxed{-36330}