Question

phân tích đa thức thành nhân tử

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

Answer 1

=a³(b-c)-b³(a-c)+c³(a-c-b+c)

=a³(b-c)-b³(a-c)+c³(a-c)-c³(b-c)

=(a³-c³)(b-c)-(b³-c³)(a-c)

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

=(b-c)(a-c)(a²+ac+c²-b²-bc-c²)

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

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

Answer 2

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

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

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

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

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

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

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

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