Question

Phân tích thành nhân tử:

a, \(a\left(b-c\right)^3+b\left(c-a\right)^3+c\left(a-b\right)^3\)

b, \(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

Answer 1

a. \(a\left(b-c\right)^3+b\left(c-a\right)^3+c\left(a-b\right)^3\)

Viết c - a dưới dạng \(-[\left(b-c\right)+\left(a-b\right)]\) , ta được:

\(A=a\left(b-c\right)^3-b[\left(b-c\right)+\left(a-b\right)]^3+c\left(a-b\right)^3\)

Áp dụng công thức \(\left(x+y\right)^3=x^3+3xy\left(x+y\right)+y^3\) , ta được:

\(A=a\left(b-c\right)^3-b[\left(b-c\right)^3+3\left(b-c\right)\left(a-b\right)\left(a-c\right)+\left(a-b\right)^3]+c\left(a-b\right)^3\)

\(=\left(b-c\right)^3\left(a-b\right)-3b\left(b-c\right)\left(a-b\right)\left(a-c\right)-\left(a-b\right)^3\left(b-c\right)\)

\(=\left(b-c\right)\left(a-b\right)[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2]\)

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

Answer 2

b. \(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)-b^4[\left(b-c\right)+\left(a-b\right)]+c^4\left(a-b\right)\)

\(=\left(b-c\right)\left(a^4-b^4\right)-\left(a-b\right)\left(b^4-c^4\right)\)

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

\(=\left(b-c\right)\left(a-b\right)\left(a^3+ab^2+a^2b+b^3-b^3-bc^2-b^2c-c^3\right)\)

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