己知方程x2－x－1＝0的根是方程x6－px2＋q＝0的根,则p＝________,q＝________.

设方程x²-x-1=0的二根分别为x₁、x₂,由韦达定理,得：
x₁+x₂=1
x₁*x₂=-1
则：
x₁²+x₂²=(x₁+x₂)²-2x₁*x₂=1+2=3
(x₁²)²+(x₂²)²
=(x₁²+x₂²)²-2x₁²*x₂²
=(x₁²+x₂²)²-2(x₁*x₂)²
=3²-2*(-1)²
=7
将x₁、x₂分别代入方程x^6－px²＋q＝0得：
x₁^6－px₁²＋q＝0······①
x₂^6－px₂²＋q＝0······②
①-②,得：
(x₁^6-x₂^6)-p(x₁²-x₂²)=0
[(x₁²)³-(x₂²)³]-p(x₁²-x₂²)=0
(x₁²-x₂²)[(x₁²)²+(x₂²)²+x₁²*x₂²]-p(x₁²-x₂²)=0
由于x₁≠x₂,则x₁²-x₂²≠0,所以化简得：
[(x₁²)²+(x₂²)²+x₁²*x₂²]-p=0
则：
p=(x₁²)²+(x₂²)²+(x₁*x₂)²
=7+(-1)²
=8
①+②,得：
(x₁^6+x₂^6)-8(x₁²+x₂²)+2q=0
[(x₁²)³+(x₂²)³]-24+2q=0
(x₁²+x₂²)[(x₁²)²+(x₂²)²-x₁²*x₂²]-24+2q=0
3[(x₁²)²+(x₂²)²-(x₁*x₂)²]-24+2q=0
3(7-1)-24+2q=0
得：q=3；
综上,得：p=8,q=3.