1.若多项式x的4次方-3x的3次方+3x的2次方+mx+n有一个因式是x的平方+1,求m,n的值.然后对原来的式子进行

x^4-3x^3+3x^2+mx+n
=x^4+x^2-3x^3-3x+2x^2+2+(m+3)x+n-2
=(x^2+1)x^2-3x(x^2+1)+2(x^2+1)+(m+3)x+n-2
=(x^2+1)(x^2-3x+2)+(m+3)x+n-2
所以 m+3=0,n-2=0
所以 m=-3,n=2
原式＝(x^2+1)(x-2)(x-1)
x^3+y^3
=(x+y)(x^2-xy+y^2)
=(x+y)(x^2+2xy+y^2-3xy)
=(x+y)[(x+y)^2-3xy]
=1×[1^2-3×(-1)]
=4