Factor Theorem problem?

Let (n - d), n and (n + d) be the three real roots of the equation.

Plug separately (n - d), n and (n + d) into the equation x³ + ax² + bx + c = 0:
(n - d)³ + a(n - d)² + b(n - d) + c = 0 …… {1}
n³ + an² + bn + c = 0 …… {2}
(n + d)³ + a(n + d)² + b(n + d) + c = 0 …… {3}

{2} - {1}:
[n³ - (n - d)³] + a[n² - (n - d)²] + b[n - (n - d)] = 0
(n³ - n³ + 3n²d - 3nd² + d³) + a(n² - n² + 2nd - d²) + b(n - n + d) = 0
(3n²d - 3nd² + d³) + a(2nd - d²) + bd = 0 …… {4}

{3} - {2}:
[(n + d)³ - n³] + a[(n + d)² - n²] + b[(n + d) - n] = 0
(n³ + 3n²d + 3nd² + d³ - n³) + a(n² + 2nd + d² - n²) + b(n + d - n) = 0
(3n²d + 3nd² + d³) + a(2nd + d²) + bd = 0 …… {5}

{5} - {4}:
[(3n²d + 3nd² + d³) - (3n²d - 3nd² + d³)] + a[(2nd + d²) - (2nd - d²)] + (bd - bd) = 0
6nd² + 2d²a = 0
a = -3n

Plug a = -3n into {5}:
(3n²d + 3nd² + d³) + (-3n)(2nd + d²) + bd = 0
3n²d + 3nd² + d³ - 6n²d - 3nd² + bd = 0
bd = 3n²d - d³
b = 3n² - d²

Plug a = -3n and b = 3n² - d into {2}:
n³ + (-3n)n² + (3n² - d²)n + c = 0
n³ - 3n³ + 3n³ - d²n + c = 0
c = -n³ + d²n


Prove that: 2a³ + 27c = 9ab

L.H.S.
= 2a³ + 27c
= 2(-3n)³ + 27(-n³ + d²n)
= -54n³ - 27n³ + 27d²n
= -81n³ + 27d²n
= -27n(3n² - d²)
= 9(-3n)(3n² - d²)
= 9ab
= R.H.S.

Hence, 2a³ + 27c = 9ab

Suppose the first term of the progression is m and the common difference is n, so the second term is m+n and the third is m+2n. So the equation is:
(x - m)(x - (m+n))(x - (m+2n)) = 0
(x - m)(x - m - n)(x - m - 2n) = 0
x^3 + (-3m-3n)x^2 + (2n^2+6mn+3m^2)x + (-m^3-3m^2n-2mn^2) = 0

So we have:
a = -3m - 3n
b = 2n^2 + 6mn + 3m^2
c = -m^3 - 3m^2n - 2mn^2

So let's see if this is true:
2a^3 + 27c = 9ab
2(-3m - 3n)^3 + 27(-m^3 - 3m^2n - 2mn^2) = 9(-3m - 3n)(2n^2 + 6mn + 3m^2)
-81m^3 - 243m^2n - 216mn^2 - 54n^3 = -81m^3 - 243m^2n - 216mn^2 - 54n^3
So it's true!

x³ + ax^2 + bx + c = 0

let the roots be u,v,w
in AP:
u = k-d, v = k, w = k+d

as roots:
(x-u)(x-v)(x-w) = 0
x³ + ax^2 + bx + c = 0

a = -(u+v+w) = -3k
b = uv+uw+vw 
= k(k-d)+(k+d)(k-d)+k(k+d)
= 3k²-d²
c = -uvw = -k(k²-d²) = -k³ + kd²

Prove that: 
2a³ + 27c = 9ab

LHS: 2a³ + 27c
-54k³ -27k³ +27kd²
= -81k³ + 27kd²

RHS: 9ab
-27k(3k²-d²)
= -81k³ + 27kd²

QED

(x - (m - d)) * (x - m) * (x - (m + d)) = x^3 + ax^2 + bx + c
(x^2 - x * (m + d) - x * (m - d) + (m^2 - d^2)) * (x - m) = x^3 + ax^2 + bx + c
(x^2 - x * (m + d + m - d) + (m^2 - d^2)) * (x - m) = x^3 + ax^2 + bx + c
(x^2 - 2mx + (m^2 - d^2)) * (x - m) = x^3 + ax^2 + bx + c
x^3 - mx^2 - 2mx^2 + 2m^2 * x + (m^2 - d^2) * x - m * (m^2 - d^2) = x^3 + ax^2 + bx + c
x^3 - 3mx^2 + (2m^2 + m^2 - d^2) * x + (md^2 - m^3) = x^3 + ax^2 + bx + c
x^3 - 3mx^2 + (3m^2 - d^2) * x + (md^2 - m^3) = x^3 + ax^2 + bx + c

-3m = a
3m^2 - d^2 = b
md^2 - m^3 = c

2a^3 + 27c = 9ab
2 * (-3m)^3 + 27 * (md^2 - m^3) = 9 * (-3m) * (3m^2 - d^2)
2 * (-27m^3) + 27md^2 - 27m^3 = -27m * (3m^2 - d^2)
-54m^3 - 27m^3 + 27md^2 = -81m^3 + 27md^2
-81m^3 + 27md^2 = -81m^3 + 27md^2
0 = 0
Confirmed.