Question

1,已知X²-x-1=0,求（x+1）³+（x²+x-1）²-12x的值?2,已

Answer 1

1.展开所求的式子,合并同类项,得,x4+3x3+2x2-11x+2,将此式改写为,(x4-x3-x2)+(4x3-4x2-4x)+(7 x2-7x-7)+9,7由于x2-x-1=0,所以易知此式的值为9

2.因为(a+b+c)3=a3+b3+c3+(3ab2+3ac2)+(3bc2+3a2b)+(3a2c+3b2c)+6abc

= (a3+b3+c3)+3a(1-a2)+ 3b(1-b2)+3c(1-c2)+6abc=3(a+b+c)-2(a3+b3+c3)+6abc,

由已知条件,易知abc=0