tk mình đi mình giải cho 

\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)

\(=\left(a^2b-b^2a\right)-\left(a^2c-b^2c\right)+c^2\left(a-b\right)\)

\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)\)

\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

=a²b-a²c+b²c-ab²+ac²-bc²

=ab(a-b)-c(a-b)(a+b)+c²(a-b)

=(a-b)(ab-c+c²)

a(b² - c²) + b(c²-a²) + c(a²-b²)

=a(b²-c²) + b(a²-b²+b²-c²) + c(a²-b²)

=a(b²-c²) - b[(a²-b²)+(b²-c²)] + c(a²-b²)

=a(b²-c²) - b(a²-b²) - b(b²-c²) + c(a²-b²)

=[a(b²-c²) - b(b²-c²)] - [b(a²-b²)-c(a²-b²) ]

=(b²-c²)(a-b) - (a²-b²)(b-c)

=(a-b)(b-c)(b+c) - (a+b)(a-b)(b-c)

=(a-b)(b-c)[(b+c)-(a+b)]

=(a-b)(b-c)(c-a)

\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=ab^2-ac^2+bc^2-ba^2+c\left(a-b\right)\left(a+b\right)\)

\(=ab^2-a^2b+bc^2-ac^2+c\left(a-b\right)\left(a+b\right)\)

\(=ab\left(b-a\right)+c^2\left(b-a\right)-c\left(b-a\right)\left(b+a\right)\)

=(b-a)\(\left(ab+c^2-cb-ca\right)\)

=(b-a)[a(b-c)+c(c-b)]

=(b-a)[a(b-c)-c(b-c)]

=(b-a)(b-c)(a-c)

(b-a)*(c-a)*(c-b)*(c+b+a)

a(b+c)^2(b-c)+b(c+a)^2(C-a)+c(a+b)^2(a-b)

=a(b+c)^2(b-a)+bc(c+a)^2-ab(c+a)^2+ac(a+b^2-cb(a+b)^2

=a(b+c)^2(b-c)+cd(c-d)(2a+b+c)-a(bc^2+ba^2-ca^2-cb^2)

=(b-c)(ab^2+ac^2+2abc-2abc-cb^2-c^2b)a(b-c)(a^2-cb)

=(b-c)(ab^2+ac^2-ab^2-c^2b-a^3+abc)

=(b-c)(a-c)[b^2-(ac+a^2)+cb]

=(b-c)(a-c)(b-a)(a+b+c)