多項式問題，請問如何解題?

方法一：
2x(x - 1) + ax(x + 1) - (x + 1)(x - 1) = 2x(x - 2) + b(x - 2) + c
2x² - 2x + ax² + ax - x² + 1 = 2x² - 4x + bx - 2b + c
(a + 1)² + (a - 2)x + 1 = 2x² + (b - 4)x + (c - 2b)

比較兩邊 x² 項的係數：
a + 1 = 2
a = 1

比較兩邊 x 項的係數：
a - 2 = b - 4
1 - 2 = b - 4
b = 3

比較兩邊常數項：
c - 2b = 1
c - 2(3) = 1
c = 7

a + b + c = 1 + 3 + 7 = 11


方法二：

2x(x - 1) + ax(x + 1) - (x + 1)(x - 1) = 2x(x - 2) + b(x - 2) + c

當 x = 1 :
a(1)(1 + 1) = 2(1)(1 - 2) + b(1 - 2) + c
2a = -2 - b + c
2a + b - c = -2 ...... [1]

當 x = -1 :
2(-1)(-1 - 1) = 2(-1)(-1 - 2) + b(-1 - 2) + c
4 = 6 - 3b + c
3b = c + 2
b = (c + 2)/3 ...... [2]

當 x = 2 :
2(2)(2 - 1) + a(2)(2 + 1) - (2 + 1)(2 - 1) = c
4 + 6a - 3 = c
6a = c - 1
a = (c - 1)/6 ...... [3]

將 [2] 及 [3] 代入 [1] 中：
2[(c - 1)/6] + [(c + 2)/3] - c = -2
[(c - 1)/3] + [(c + 2)/3] - c = -2
(c - 1) + (c + 2) - 3c = -6
c - 1 + c + 2 - 3c = -6
c = 7

將 c = 7 代入 [2] 中：
b = (7 + 2)/3
b = 3

將 c = 7 代入 [3] 中：
a = (7 - 1)/6
a = 1

a + b + c = 1 + 3 + 7 = 11