Question

phân tích đa thức thành nhân tử

a, ab(a + b) + bc(b + c) + ca(c+a) + 3abc

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

c, x³ + 3x² + 6x + 18

Answer 1

a, ab(a + b) + bc(b + c) + ca(c+a) + 3abc

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

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

Answer 2

b)Đề bài=

\(a^2b+ab^2-b^2c-bc^2+a^2c-ac^2+abc-abc\)

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

Answer 3

c)\(x^3+3x^2+6x+18\)

=\(x^2\left(x+3\right)+6\left(x+3\right)=\left(x+3\right)\left(x^2+6\right)\)