Maths Questions - F.3

因為 (a-b)^2=(b-a)^2 及 (a-b)^3=-(b-a)^3, 所以
2(a-b)^2 - (a-b)^3 + (b-a)
= 2(b-a)^2 + (b-a)^3 + (b-a)
= (b-a)[2(b-a) + (b-a)^2 +1]
= (b-a)[(b-a)^2 + 2(b-a) +1]
= (b-a)(b-a+1)^2

or -(a-b)(a-b-1)^2
因為 (b-a+1)^2=(a-b-1)^2, 樓上那個漏了個負號。

2(a-b)-(a-b)^3+(b+a)=1

2(a-b)^2 - (a-b)^3 + (b-a)
2(a-b)^2 - (a-b)^3 - (-b+a)
2(a-b)^2 - (a-b)^3 - (a - b)
2(a-b)^2 - (a-b)^3 - (a - b)
(a-b)^3 -2 (a-b)^2 + (a - b)
(a – 1) is the common factor

(a – b)[(a – b)^2 – 2(a-b) +1]
Note: x^2 – 2x + 1 = (x – 1)^2
(a – b)[(a – b) - 1] [(a – b) - 1]
(a – b)[a – b - 1] [a – b - 1]
(a – b)(a – b - 1 )^2

2(a-b)^2 - (a-b)^3 + (b-a) = (a – b)(a – b - 1 )^2 (Answer)