Question

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

Answer 1

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