請教以下三題因式分解,謝謝?

1.
令 a = y-z , b = z-x , c = x-y , 則 a+b+c=0.
原式 = a⁵ + b⁵ + c⁵
= (a + b)(a⁴- a³b + a²b² - ab³ + b⁴) - (a + b)⁵
= (a + b)(a⁴- a³b + a²b² - ab³ + b⁴- (a + b)⁴)
= (a + b)(a⁴- a³b + a²b² - ab³ + b⁴- a⁴- 4a³b - 6a²b² - 4ab³ - b⁴)
= (a + b)( - 5a³b - 5a²b² - 5ab³)
= - 5(a + b)ab(a² + ab + b²)
= 5abc(a² + ab + b²)
= 5abc(a² + b² + a²+2ab+b²)/2
= 5abc(a² + b² + c²)/2
= 5(y - z)(z - x)(x - y)(x² + y² + z² - xy - yz - zx)

2.
(y+z)(z+x)(x+y) + xyz
= (y+z)(x²+(y+z)x+yz) + xyz
= (y+z)x² + ((y+z)²+yz)x + (y+z)yz
視之為x的二次多項式,十字交乘得
= ((y+z)x + yz) (x + y+z)
= (x+y+z)(xy+yz+zx)

3.
(x+y+z)³ - (y+z-x)³ - (z+x-y)³ - (x+y-z)³
= (x + y+z)³ + (x - (y+z))³ - [(x - (y-z))³ + (x + y-z)³]
= 2x³ + 6x(y+z)² - [2x³ + 6x(y-z)²]
= 6x [(y+z)² - (y-z)²]
= 6x (4yz)
= 24xyz