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a) =a2b - ab2 + b2c - bc2 + a2c - ac2
= abc +a2b - ab2 +b2c - bc2 +a2c - ac2 - abc
= (a2b - abc) - (ab2 - b2c) - (bc2 - ac2) - (a2c - abc)
= ab(a - c) - b2(a - c) - c2(b - a) - ac(a - b)
= [ab(a - c) - b2(a - c)] + [c2(a - b) - ac(a - b)]
= (a - c)(ab - b2) + (a - b)(c2 - ac)
= b(a - c)(a - b) + c(a - b)(c - a)
= b(a - c)(a - b) - c(a - b)(a - c)
= (a - c)(a - b)(b - c)
b)= ab2 - ac2 + bc2 - a2b + a2c - b2c
= abc + ab2 - ac2 + bc2 - a2b + a2c - b2c - abc
= (ab2 - abc) + (abc - ac2) - (b2c - bc2) - (a2b - a2c)
= ab(b - c) + ac( b - c) - bc(b - c) - a2(b - c)
= (b - c)(ab + ac - bc - a2)
= (b - c) [(ab - bc) + (ac - a2)]
= (b - c) [b(a - c) +a(c - a)]
= (b - c) [b(a - c) - a(a - c)]
= (b - c)(a - c)(b - a)
c) = ab3 - ac3 + bc3 - a3b + a3c - b3c
= a2bc + ab2c + abc2 + a3b + a2b2 + a2bc - a3c - a2bc - a2c2 + a2c2 + abc2 + ac3 - a2b2
- ab3 - ab2c + ab2c + b3c + b2c2 - abc2 - b2c2 - bc3 - a2bc - ab2c - abc2
= (a2bc + ab2c + abc2) +(a3b + a2b2 + a2bc) - (a3c - a2bc - a2c2) +(a2c2 + abc2 +ac3) -
(a2b2 + ab3 + ab2c) + (ab2c + b3c + b2c2) - (abc2 + b2c2 + bc3) - (a2bc + ab2c + abc2)
= abc(a + b + c) +a2b(a + b + c) - a2c(a + b + c) + ac2(a + b + c) - ab2(a + b + c) + b2c(a + b + c) - bc2(a + b + c) - abc(a + b+ c)
= (a +b +c)(abc + a2b - a2c + ac2 - ab2 + b2c - bc2 - abc)
= (a + b+ c) [(a2b - abc)+(abc - bc2) - (a2c - ac2) - (ab2 - b2c)]
= (a + b + c) [ab(a - c) + bc(a - c) - ac(a - c) - b2(a - c)]
= (a + b + c)(a - c)(ab + bc - ac - b2)
= (a +b + c)(a - c) [(ab - ac) - (b2 - bc)]
= (a + b+ c)(a - c) [a(b - c) - b(b - c)]
= (a + b + c)(a - c)(b - c)(a - b)
trời ơi sao câu c dài thế !!!!! Tui có bài giống vậy nhưng nó ra p/số, còn phải ghi nhiều hơn 
nhân tung \(\left(a^2-b\right)\left(b^2-c\right)\left(c^2-a\right)\) ra đề rồi viết ngược lại =.=
a)\(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a-c\right)\)
b)\((a+b)(a^2-b^2)+(b+c)(b^2-c^2)+(c+a)(c^2-a^2)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
c)\(a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ca\right)\)
d)\(a^4(b-c)+b^4(c-a)+c^4(a-b)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2+ab+bc+ca\right)\)
a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2\left[\left(c-b\right)-\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)\)
c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\left(1\right)\)
Đặt \(x^2+8x+11=y\)Thay vào (1) ta được
\(\left(y-4\right)\left(y+4\right)+15\)
\(=y^2-16+15\)
\(=y^2-1\)
\(=\left(y-1\right)\left(y+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+11\right)\)
=a(b + c)^2(b - c) + b(c + a)^2(c - a) - c(a + b)^2[(b - c) + (c - a)]
= a(b + c)^2(b - c) + b(c + a)^2(c - a) - c(a + b)^2(b - c) - c(a + b)^2(c - a)
= [a(b + c)^2(b - c) - c(a + b)^2(b - c)]+ [b(c + a)^2(c - a) - c(a + b)^2(c - a)]
= (b - c)[a(b + c)^2 - c(a + b)^2] + (c - a)[b(c + a)^2 - c(a + b)^2]
= (b - c)(ab^2 + ac^2 - ca^2 - cb^2) + (c - a)(bc^2 + ba^2 - ca^2 - cb^2)
= (b - c)[ac(c - a) - b^2(c - a)] + (c - a)[a^2(b - c) - bc(b - c)]
= (b - c)(c - a)(ac - b^2) + (c - a)(b - c)(a^2 - bc)
= (b - c)(c - a)(ac - b^2 + a^2 - bc)
= (b - c)(c - a)[(a^2 - b^2) + (ac - bc)]
= (b - c)(c - a)[(a - b)(a + b) + c(a - b)]
= (b - c)(c - a)(a - b)(a + b + c)
= (a - b)(b - c)(c - a)(a + b + c).