c^4-4c^3+4c^2-1=0,c=?

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c^4-4c^3+4c^2-1=0,c=?
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c^4-4c^3+4c^2-1=0,c=?
c^4-4c^3+4c^2-1=0,c=?

c^4-4c^3+4c^2-1=0,c=?
c^4-4c^3+4c^2-1
=c^4-c^3-3c^3+3c^2+c^2-1
=c^3(c-1)-3c^2(c-1)+(c-1)(c+1)
=(c-1)(c^3-3c^2+c+1)
=(c-1)(c^3-c^2-c^2+c-c^2+1)
=(c-1)[c^2(c-1)-c(c-1)-(c-1)(c+1)]
=(c-1)^2(c^2-2c-1)=0
c=1,1+√2,1-√2.

显然c=1是一解
所以c^4-4c^3+4c^2-1一定有c-1的因式
c^4-4c^3+4c^2-1=(c-1)(c^3-3c^2+c+1)
同理c^3-3c^2+c+1也c-1的因式
c^3-3c^2+c+1=(c-1)(c^2-2c-1)
c^2-2c-1=0的根为±√2 +1
c^4-4c^3+4c^2-1=0的根为 c=1 c=√2 +1 c=-√2 +1