Question

Phân tích đa thức thành nhân tử

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

Answer 1

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

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

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

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

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

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

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

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

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

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

Answer 2

Cách khác:

Ta có:
\(a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)\)

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

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

\(=(a-b)(b^2-c^2)-(b-c)(a^2-b^2)\)

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

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