a)\(x^3+y^3+z^3-3xyz\)
\(=x^3+3x^2y+3xy^2+y^3+z^3-3xyz-3x^2y-3xy^2\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\)
b) x4 + 4
=x4+4x2+4-4x2
=(x2+2)2-(2x)2
=(x2+2-2x)(x2+2+2x)
c)sai đề
c) x^3 - x + 3x^2y + 3xy^2 + y^3 - y
=(x^3 + 3x^2y + 3xy^2 + y^3) - ( x + y )
=(x+y)^3 - (x+y)
=(x+y)(x^2+2xy+y^2-1) = (x+y)(x+y-1)(x+y+1)
b) \(x^4+4=\left(x^2\right)^2+2^2=\left(x^2-2\right)\left(x^2+2\right)\)
a)Ta có:
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).
