设方程x^2+3x+1=0的两根为x1,x2,求x1^7+x2^7的值!

x1+x2=-3
x1x2=1
则x1²+x2²=(x1+x2)²-2x1x2=7
平方
x1^4+x2^4+2x1²x2²=49
x1^4+x2^4=49-2(x1x2)²=47
x1³+x2³
=(x1+x2)(x1²-x1x2+x2²)
=-3*(7-1)
=-18
(x1^4+x2^4)(x1³+x2³)=47*(-18)
x1^7+x2^7+x1^4x2³+x1³x2^4=-846
x1^7+x2^7+(x1x2)³(x1+x2)=-846
x1^7+x2^7+1³*(-3)=-846
x1^7+x2^7=-843