初二数学----->分解因式..(9)

设a^2+b^2=x>0
则
x(x-1)=20
x^2-x-20=0
(x-5)(x+4)=0
因为x>0
x+4>4
所以
x-5=0
x=5
既 a^2+b^2=5
2.x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)
因为x+y+z=0
所以
x^3+y^3+z^3-3xyz=0
既x^3+y^3+z^3=3xyz

1.(a^2+b^2)(a^2+b^2-1)-20=(a^2+b^2)^2-(a^2+b^2)-20
=(a^2+b^2+4)(a^2+b^2-5)
2.x^3+y^3+z^3=(x+y)(x^2-xy+y^2)+z^3=(-z)(x^2+2xy+y^2-3xy)+z^3
=(-z)(x+y)^2-3xy(-z)+z^3=-z(-z)^2+3xyz+z^3=3xyz
得证

设a^2+b^2=x
x(x-1)=20
a^2+b^2>0
x=5

x^3+y^3=(x+y)(x^2-xy+y^2)
x+y=-z x^2+y^2=z^2-2xy
把括号内的东西相乘
代入可得x^3+y^3+z^3=3xyz