Question

Rút gọn phân thức : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}.\)

Answer 1

Phân tích mẫu thức thành nhân tử :

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

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

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

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

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

Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)

Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)

Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)

Answer 2

Tử:

(b - c)³ + (c - a)³ + (a - b)³

= (b - c + c - a + a - b)³ - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)

= 0 - 3(b - a)(a - c)(c - b)

= 3(a - b)(a - c)(c - b)

Mẫu:

a²(b - c) + b²(c - a) + c²(a - b)

= a²(b - c) + b²c - ab² + ac² - bc²

= a²(b - c) - a(b² - c²) + bc(b - c)

= a²(b - c) - a(b - c)(b + c) + bc(b - c)

= (b - c)(a² - ab - ac + bc)

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

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

\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)

\(=\frac{3\left(c-b\right)}{b-c}\)