x3-y3-z3等于3xyz,x2等于7(y-z)

根据题意得：（x3-2y3）+（x3+y3）=x3-2y3+x3+y3=2x3-y3．故答案为：2x3-y3．

a^3+b^3=(a+b)(a^2-ab+b^2)a^3-b^3=(a-b)(a^2+ab+b^2)(x+y+z)^3-x^3-y^3-z^3=(y+z)[(x+y+z)^2+(x+y+z)x+x^2

首先这三个点肯定可以确定一个平面,还可以确定一条到这三个点距离都相等的直线.那么,一个平面和一条直线的交点,就一定是唯一的圆心.我现在的想法是：1、设三点确定的平面是z=a1x+b1y+c1,带入三个

（1）原式=-2xn-1yn（x4n-2x2ny2+y4）=-2xn-1yn[（x2n）2-2x2ny2+（y2）2]=-2xn-1yn（x2n-y2）2=-2xn-1yn（xn-y）2（xn+y）2

x+y+z=0,z=-（x+y）代入x^3+y^3+z^3=0x^3+y^3-(x+y)^3=x^3+y^3-x^3-3X^2y-3xy^2-y^3=-3xy(x+y)=3xyz=0所以xyz=03a

∵x+y+z=0,∴z=(-x-y)x^3+y^3+z^3=x^3+y^3-(x+y)^3=x^3+y^3-x^3-y^3-3x^2y-3xy^2=-3xy(x+y)=3xyz

因为:X3-Y3-Z3=3XYZ所以:X3+(-Y)3+(-Z)3-3X(-Y)(-Z)=0(X-Y-Z)(X2+Y2+Z2+XY+XZ-YZ)=0所以:1.X-Y-Z=02.X2+Y2+Z2+XY+

由(x+y+z)2-(x2+y2+z2)可得xy+xz+yz=-5x3+y3+z3-3xyz=（x+y+z)(x2+y2+z2-xy-yz-zx）可得xyz=14再问：谢谢，我看一下其他的答案在采纳再

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（x+y+z）²-（x²+y²+z²）=2（xy+yz+zx）=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&

∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2

(x+y+z)^3-x^3-y^3-z^3=3yx^2+3xy^2+3xz^2+3yz^2+3zx^2+3zy^2+6xyz=3xy(x+y)+3z^2(x+y)+3z(x^2+y^2+2xy)=3x

如果你的X2是x的平方,X3是x的三次方那么答案是：-(x-y+z)*(x-y-z)*(x+y-z)

下面是我的思路,尽量用Matlab语言叙述的,方便你作图.假设：(x1,y1,z1),(x2,y2,z2),(x3,y3,z3),(x0,y0,z0),R,(A,B,C,D)均已知.法向量(A,B,C

x3+y3-z3+3xyz，=[（x+y）3-3x2y-3xy2]-z3+3xyz，=[（x+y）3-z3]-（3x2y+3xy2-3xyz），=（x+y-z）[（x+y）2+（x+y）z+z2]-3

x3（y-z）+y3（z-x）+z3（x-y）=x3（y-z）+y3（z-x）-z3（y-z）-z3（z-x）=（x3-z3）（y-z）+（y3-z3）（z-x）=（x-z）（y-z）（x2+xz+z

a+b+c)(ab+bc+ca)-abc=a^2b+2abc+ca^2+ab^2+b^2c+bc^2+c^2a=(a^2b+ab^2)+(bc^2+ac^2)+(2cab+ca^2+cb^2)=ab(

x^3+y^3+z^3+(x+y)^3+(y+z)^3+(z+x)^3=[x^3+(y+z)^3]+[y^3+(z+x)^3]+[z^3+(x+y)^3]=(x+y+z)(x^2-xy+y^2-xz+

因为x^3+y^3+z^3-3xyz=(x+y)^3-3x^y-3xy^2+z^3-3xyz（把x^3+y^3写成(x+y)^3-3x^2y-3xy^2）=[(x+y)^3+z^3]-(3x^2y+3

x^3+y^3+z^3-3xyz=[(x+y)^3-3x^2y-3xy^2]+z^3-3xyz=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+