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

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

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

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

nhân hả bạn

ab(b - a) - bc(b - c) - ac(c - a)

= ab² - a²b - b²c +  bc² + ac(a - c)

= b²(a - c) - b(a² - c²) + ac(a - c)

= b²(a - c) - b(a - c)(a + c) + ac(a - c)

= (b² - ab - bc + ac)(a - c)

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

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

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)

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

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

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

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

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

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

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

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

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)

\(=ab\left[\left(b-c\right)+\left(c-a\right)\right]-bc\left(b-c\right)-ac\left(c-a\right)\)

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

\(=\left[ab\left(b-c\right)-bc\left(b-c\right)\right]+\left[ab\left(c-a\right)-ac\left(c-a\right)\right]\)

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

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

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

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

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

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

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

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

Ta có b + c = (a + b) + (c – a) nên

A = ab(a + b) – bc[(a + b) + (c – a)] – ac(c – a)

= ab(a + b) – bc(a + b) – bc(c – a) – ac(c – a)

= b(a + b)(a – c) – c(c – a)(b + a)

= (a + b)(a – c)(b + c)

Đáp án cần chọn là: B