x^4﹢64 分解因式

补充（分解因式）x³﹣11x²﹢31x﹣21

                            x³﹣4xy²﹣2x²y﹢8y³

已知a﹢b﹢c＝0,求证：a³﹢a²c﹢b²c﹣abc﹢b³＝0

已知实数x、y满足x²﹢y²﹣xy﹢2x﹣y﹢1＝0,试求x、y的值

请注明题号，加过程与讲解。由于当前的题目过多，所以请写规范，谢谢。

1、x^3-11x^2+31x-21      =x^3-x^2-10x^2+10x+21x-21     =x^2(x-1)-10x(x-1)+21(x-1)     =(x-1)(x^2-10x+21)     =(x-1)(x-3)(x-7) 2、 x³﹣4xy²﹣2x²y﹢8y³    =x³﹣2x²y﹢8y³﹣4xy²    =x^2(x-2y)-4y^2(x-2y)    =(x^2-4y^2)(x-2y)    =(x+2y)(x-2y)(x-2y)    =(x+2y)(x-2y)^2

x^4+64=x^4+16x^2+64-16x^2=(x^2+8)^2-16x^2=(x^2+8-4x)(x^2+8+4x) x^3-11x^2+31x-21=x^3-x^2-10x^2+10x+21x-21=x^2(x-1)-10x(x-1)+21(x-1) =(x-1)(x^2-10x+21)=(x-1)(x-3)(x-7) a^3+a^2c+b^2c-abc+b^3=a^2(a+c)+b^2(b+c)-abc= (a^2)(-b)+(b^2)(-a)-abc=-ab(a+b+c)=0 x^2+y^2-xy+2x-y+1=0 x^2+2x+1-y(x+1)+y^2=0 (x+1)^2-y(x+1)+y^2=0 (x+1-y/2)^2+3y^2/4=0 则y=0,x+1-y/2=0,x=-1 即x=-1,y=0