分解因式:(b+c-2a)^3+(c+a-2b)^3+(a+b-2c)^3
具体过程
人气:238 ℃ 时间:2019-09-16 20:38:47
解答
设a-b=d,b-c=e,c-a=f
(b+c-2a)^3+(c+a-2b)^3+(a+b-2c)^3
=(f-d)^3+(d-e)^3+(e-f)^3
=(f-e)((f-d)^2-(f-d)(d-e)+(d-e)^2)-(f-e)^3
=(f-e)((f-d)^2-(f-d)(d-e)+(d-e)^2-(f-e)^2)
=(f-e)((f-d)(f-d-d+e)+(d-e-f+e)(d-e+f-e))
=(f-e)((f-d)(f+e-2d)-(f-d)(f+d-2e))
=(f-e)(f-d)(f+e-2d-f-d+2e)
=(f-e)(f-d)(3e-3d)
=3(f-e)(f-d)(e-d)
=3(c-a-(b-c))(c-a-(a-b))(b-c-(a-b))
=3(2c-a-b)(b+c-2a)(2b-c-a)
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