Question

已知x+y=-2,a+b=-1/2,x2-xy+y2=-1,求bx3+ay3+by3+ax3的值

Answer 1

bx3+ay3+by3+ax3
=x³(b+a)+y³(a+b)
=(a+b)(x³+y³)
=(a+b)(x+y)(x²-xy+y²)
=(-1/2)*(-2)*(-1)
=-1

Answer 2

答案是-1
解：bx3+ay3+by3+ax3 注明：6xy=2<(x+y)2-(x2-xy+y2)
=(a+b)(x3+y3) 化简 =10
=-1/2(x3+y3) 代入
=-1/2<(x+y)3+6xy> x3+y3=<(x+y)3+6xy> 此步也是经化简得到的
=-1/2(-8+10)
=-1