已知二次函数f(x)=ax^2+bx+c满足|f(1)|=|f(-1)|=|f(0)|=1,求f(x)的表达式

|a+b+c|=1
|a-b+c|=1
|c|=1

(1)
a+b+c=1
a-b+c=1
c=1
=> a=0, b=0, c=1
f(x) = 1

(2)
a+b+c=-1
a-b+c=1
c=1
=> a=-1, b=-1, c=1
f(x) = -x²-x+1

(3)
a+b+c=1
a-b+c=-1
c=1
=> a=-1, b=1, c=1
f(x) = -x²+x+1

(4)
a+b+c=1
a-b+c=1
c=-1
=> a=2, b=0, c=-1
f(x) = 2x²-1

(5)
a+b+c=-1
a-b+c=-1
c=1
=> a=-2, b=0, c=1
f(x) = -2x²+1

(6)
a+b+c=-1
a-b+c=1
c=-1
=> a=1, b=-1, c=-1
f(x) = x²-x-1

(7)
a+b+c=1
a-b+c=-1
c=-1
=> a=1, b=1, c=-1
f(x) = x²+x-1

(8)
a+b+c=-1
a-b+c=-1
c=-1
=> a=0, b=0, c=-1
f(x) = -1

所以共有8个方程符合条件：
f(x) = 1
f(x) = -1
f(x) = 2x²-1
f(x) = -2x²+1
f(x) = x²+x-1
f(x) = x²-x-1
f(x) = -x&su