已知x-y不等于0,x^2-x=7,y^2-y=7,求x^3+y^3+x^2y+xy^2的值
问题描述:
已知x-y不等于0,x^2-x=7,y^2-y=7,求x^3+y^3+x^2y+xy^2的值
答
已知x-y≠0,x²-x=7,y²-y=7,求x³+y³+x²y+xy²的值.
经观察可知,x、y相当于是方程t²-t=7的两个不等实根,方程整理为:
t²-t-7=0
由韦达定理,得:
x+y=1
xy=-7
所以:
x³+y³+x²y+xy²
=(x³+y³)+(x²y+xy²)
=(x+y)(x²+y²-xy)+xy(x+y)
=(x+y)(x²+y²)
=x²+y²
=(x+y)²-2xy
=1-2*(-7)
=1+14
=15