a) (x  + y + z)³ - x³ - y³ - z³

= (x + y + z)³ - z³ - (x³ + y³) 

= (x + y + z - z)[(x + y + z)² + (x + y + z).z + z²) - (x + y)(x² - xy + y²)

= (x + y)(x² + y² + z² + 2xy + 2yz + 2zx + 2xz + 2yz + z² + z²) - (x + y)(x² - xy + y²)

= (x + y)(x² + y² + 3z² + 2xy + 4yz + 4zx) - (x + y)(x² - xy + y²)

= (x + y)(3z² + 3xy + 5yz + 4zx) 

b) Sửa đề x⁴ + 2010x² + 2009x + 2010

= (x⁴ + x² + 1) + (2009x² + 2009x + 2009)

= (x⁴ + 2x² + 1 - x²) + 2009(x² + x + 1)

= [(x² + 1)² - x²] + 2009(x² + x + 1)

= (x² + x + 1)(x² - x + 1) + 2009(x² + x + 1)

= (x² + x + 1)(x² - x + 2010)

Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)

Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

\(A=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Hay \(A=3\cdot2x\cdot2y\cdot2z\)

\(A=24xyz\)

a.\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)^3+z^3\right]-a^3-b^3-c^3\)

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

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

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

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

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

b.\(x^4+2010x^2+2009x+2010\)

\(=\left(x^4-x\right)+\left(2010x^2+2010x+2010\right)\)

=\(x\left(x-1\right)\left(x^2+x+1\right)+2010\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2010\right)\)

Đặt y-z=-[(x-y)+(z-x)]

Thay vào rồi cm nha bạn