"докажите тождество (x+y+z)³-x³-y³-z³=3(x+y)(x+z)(y+z)"

(x+y+z)³ - x³ - y³ - z³ = (x+y+z)(x+y+z)(x+y+z) - x³ - y³ - z³

(x+y+z)(x+y+z)(x+y+z) = x(x+y+z)(x+y+z) + y(x+y+z)(x+y+z) + z(x+y+z)(x+y+z)

= x³ + xy(x+y+z) + xz(x+y+z) + yx(x+y+z) + y²(x+y+z) + yz(x+y+z) + zx(x+y+z) + zy(x+y+z) + z²(x+y+z)

= x³ + y³ + z³ + 3xy(x+y+z) + 3xz(x+y+z) + 3yx(x+y+z) + 3y² + 3yz(x+y+z) + 3zx(x+y+z) + 3zy(x+y+z) + 3z²

= x³ + y³ + z³ + 3(x+y+z)(xy + xz + yx + y² + yz + zx + zy + z²)

= x³ + y³ + z³ + 3(x+y)(x+z)(y+z)

Теперь можно видеть, что (x+y+z)³ - x³ - y³ - z³ = 3(x+y)(x+z)(y+z)