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问题: 分解因式

1:(2+1)(2*2+1)(2*4+1)(2*8+1)
2:(1-1/2*2)(1-1/3*2)(1-1/4*2)(1-1/5*2)
3:(a-c)*2-4(b-c)(a-b)

解答:

1:(2+1)(2^2+1)(2^4+1)(2^8+1)
 = (2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
 = (2^2-1)(2^2+1)(2^4+1)(2^8+1)
 = (2^4-1)(2^4+1)(2^8+1)
 = (2^8-1)(2^8+1)
 = 2^16-1
2:(1-1/2²)(1-1/3²)(1-1/4²)(1-1/5²)
 = (1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)(1-1/5)(1+1/5)
 = (1/2)(3/2)(2/3)(4/3)(3/4)(5/4)(4/5)(6/5)
 = (1/2)(6/5)
 = 3/5
3:(a-c)²-4(b-c)(a-b)
 = [(a-b)+(b-c)]²-4(a-b)(b-c)
 = (a-b)²+(b-c)²-2(a-b)(b-c)
 = [(a-b)-(b-c)]²
 = (a+c-2b)²