Question

Phân tích đa thức thành nhân tử

a(b³-c³)+b(c³-a³)+c(a³-b³)

Answer 1

Ta có: \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

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

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

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

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

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

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

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