Question

求解：(a-b)(3a+b)^2+(a+3b)^2(b-a)

Answer 1

(a-b)(3a+b)^2+(a+3b)^2(b-a)
=(a-b)(3a+b)^2-(a+3b)^2(a-b)
=(a-b)[(3a+b)^2-(a+3b)^2]
=(a-b)[(3a+b)-(a+3b)] [(3a+b)+(a+3b)]
=(a-b)(3a+b-a-3b)(3a+b+a+3b)
=(a-b)(2a-2b)(4a+4b)
=8(a-b)(a-b)(a+b)
=8(a-b)^2(a+b)

追问

如图，已知角A=角F，角C=角D，试说明BD//CE

追答

因为角A=角F
所以AC//DF
所以角C=角CEF
因为角C=角D
所以角CEF=角D
所以BD//CE

Answer 2

=(a-b)(9a^2+6ab+b^2-a^2-6ab-9b^2)
=(a-b)(8a^2-8b^2)
=8(a-b)(a-b)(a+b)
=8(a-b)^2(a+b)