分解因式 3(x+y)^3-27(x+y) 已知x+y=-2,a+b=-1⼀2,x^2-xy+y^2=-1,求bx^2+ay^2+by^ax^2

3(x+y)^3-27(x+y)
=3(x+y)[(x+y)²-9]
=3(x+y)(x+y+3)(x+y-3)
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已知x+y=-2,a+b=-1/2,x^2-xy+y^2=-1,那么：
(x+y)²=x²+2xy+y²=-2
而xy=x²+y²+1
那么：x²+2(x²+y²+1)+y²=-2
即得：x²+y²=-4/3
所以：bx²+ay²+by²+ax²
=b(x²+y²)+a(x²+y²)
=(a+b)(x²+y²)
=(-1/2)*(-4/3)
=2/3

第一题：3（x+y）^3-27(x+y)
=3(x+y)(x^2-xy+y^2)-27(x+y)

=3(x+y)(x^2-xy+y^2-9)
因为，x+y=-2，x^2-xy+y^2=-1
所以上式=3*(-2)*(-1-9)=60
第二题你确定题目没错？

(1) 3(x+y)^3-27(x+y)=3(x+y)[(x+y)²-9]
=3(x+y)(x+y+3)(x+y-3)

(2) x+y=-2
x²-xy+y²=-1
(x+y)²-3xy=-1
xy=5/3
a+b=-1/2
bx²+ay²+by^ax²