Question

[Linh Lê]

Phân tích đa thức thành nhân tử :

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

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

\(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\)

Answer 1

[Hàn Vũ]

a(b²+c²)+b(c²+a²)+c(a²+b²)+22abc

= ab²+ac²+bc²+a²b+(a²c+b²c+2abc)

= ab(a+b)+c²(a+b)+c(a+b)²

= (a+b)(ab+c²+ac+bc)

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

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

b)

(a+b)(a²-b²)+(b+c)(b²-c²)+(a+c)(c²-a²)

= (a+b)(a²-b²)-(b+c)[(a²-b²)+(c²-a²)] +(a+c)(c²-a²)

= (a²-b²)(a+b-b-c) +(c²-a²)(a+c-b-c)

= (a²-b²)(a-c)+(c²-a²)(a-b)

= (a-b)(a²-ac+ab-bc +c²-a²)

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

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

Answer 2

[Nguyễn Ngọc Mai]

\(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\\ =a^3\left(c-b^2\right)+ab^3-b^3c^2+bc^3-a^2c^3+a^2b^2c^2-abc\\ =a^3\left(c-b^2\right)+bc^2\left(c-b^2\right)-ab\left(c-b^2\right)-a^2c^2\left(c-b^2\right)\\ =\left(c-b^2\right)\left(a^3+bc^2-ab-a^2c^2\right)\\ =\left(c-b^2\right)\left[a^2\left(a-c^2\right)-b\left(a-c^2\right)\right]\\ =\left(c-b^2\right)\left(a-c^2\right)\left(a^2-b\right)\)