Find the remainder when (x+1)^n is divided by (x-1)^3. Askiitians tutors please help. No one replied when i posted this question before. This question is of rmo. Please give the solution.

When we divide polynomial P(x) by quotient Q(x) and get remainder R(x), this means:
P(x) = D(x)*Q(x) + R(x)
Order of the polynomials: P is of order n, Q of order 3, D of order (n - 3) and R of order 2.
Let R(x) = ax^2 + bx + c:
(x + 1)^n = D(x)*(x - 1)^3 + (ax^2 + bx + c)
We can eliminate the term with unknown polynomial D(x) by substituting x = 1:
2^n = a + b + c
Differentiate both sides and substituting x = 1 again to obtain another equation:
n*[(x + 1)^(n - 1)] = g(x - 1) + (2ax + b) where g is some function of (x - 1) so when x = 1, g = 0.
n*(2^(n - 1)) = 2a + b
Differentiate original equation twice:
n*(n - 1)*[(x + 1)^(n - 2)] = h(x - 1) + 2a; h is another function where h = 0 when x = 1.
When x = 1, n*(n - 1)*(2^(n - 2)) = 2a.
Hence we have 3 equations for 3 unknowns a, b and c:
U = 2^n = a + b + c
V = n*(2^(n - 1)) = 2a + b
W = n*(n - 1)*(2^(n - 2)) = 2a
giving
a = W/2
b = V - W
c = U - (a + b) = W/2 - V + U
Then R(x) = ax^2 + bx + c
R(x) = (W/2)(x^2) + (V - W)*x + (W/2 - V + U)
To simplify this, we subsitute:
W = [2^(n - 3)]*[2n(n - 1)]
W/2 = [2^(n - 3)]*[n(n - 1)]
V = n*(2^(n - 1)) = [2^(n - 3)]*[4n]
U = 2^n = [2^(n - 3)]*[8]
R(x) = [2^(n - 3)] * [n(n - 1)(x^2) + (4n - (2n^2 - 2n))x + (n(n - 1) - 4n + 8]
R(x) = [2^(n - 3)] * [n(n - 1)(x^2) + (- 2n^2 + 6n)x + (n^2 - 5n + 8)]