Question

Phân tích đa thức sau thành nhân tử:

x³ + y³ + z³ -3xyz

Answer 1

Answer 2

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

Answer 3

\(x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(\left(x+y\right)^3+z^3\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right).z+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\)

Bài áp dụng 2 HĐT : a^3+b^3 = (a+b)^3-3ab(a+b)

(a^3+b^3)=(a+b)(a^2-ab+b^2)

Answer 4

thêm bớt 3x²y và 3xy² rồi nhóm là ra