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1、习题二5•已知/?且q+/?+c=0,求证:a5+b‘+c5a3+b3+c3a2+h2+c2=•532解.由题设a+b+c=0,推出a$+/?2+c2=+bc+ca),/+//+c3=3abc,即、10r0-(ab+be+cci)=—(tz~+b~4-(?)abc=—(«3+//+c3)3因此(左+戸+，)3+戸+。2)=a5+b‘+c5+t/2/?2(6f+/?)+/?2c2(/?+(?)+ez2c2(tz+c)=a5+b5+c5+a2b2(-c)+b2c2(-a)+a2c2(-b)=a5+b5+c5+abc^ab+be+ac)+b5+c5+a2+/异+c・•・a5+b5+c5=-(a3
2、+//+，).(/+b~+c2)6a5+/?5+c5ay+b3+c3a2+b2+c2=•5327(1)x4+y4+(x+y)°(2)x2+(x+l)2+(x2+x)2(3)(y+z)(z+x)(x+y)+xyz(4)(x-y)(x+y)3+(y-z)(y+z)3+(z—兀)(z+x)3解：(1)原式=x4+y°+(x+y)“+，)』一乳亍=x4+y4+[(兀+y)'_x)J[(兀+I')?+xy-^-x2y2=(*+尸)_兀2^2+[(乂+对2_xy][(x+)；)2+Xy]=[G2+)')+x』X+>"_x)J]+[&+y)2+xyk+)"+小)=(x24-y24-xy)[x2+y2-
3、xy+(兀+y)2+xy=(x2+xy+y2)2-(x2+y2+xy)=2(x24-xy+y2)2(2)原式=[x(x+l)]24-2x(%+1)+1=[x(兀+1)+l]2=(x2+兀+1)2(2)此多项式是对称多项式。当x=(y+z)时，/(x,”z)=(y+z)[z一(y+z)][-(y+z)+刃+[-(y+z)]yz=(y+z)(-y)(-z)(y+z)yz=0所以f(x,y,刁)有因式(x+y+z),因原式为三次式，故还有另一个二次对称式的因式，设,(y+z)(z+x)(x+y)+xyz=(x+y+z)[/n(x2+y2+z2)+n[xy+yz+xz)]令x=l,y=l,z=(
4、).得，2=(2m+n)即2m+n=1(1)令x=1,y=1,z=1.得，9=3(3m+3n)即m+n=1(2)由(1)(2)得，m=O,n=l,所以，(y+z)(z+x)(x+y)+xyz=(x+y+z)(xy+yz+zx)(4)原式是轮换多项式，当兀=y时，原式=(y-z)(y+z)‘+(z-y)(z+y)‘=0因式有因式(x-y)(y-z)(z-x),设(x一y)(x+yY+(y—z)(y+z)3+(z-x)(z+x)3=k(x+y+z)(x一y)(y-z)(z一x)令x=l.y=2,z=0得:(-l).33+2.23+(-l)l3=3R(-1)2(-1)所以k=-2・・・(x一y)
5、(x一y)3+(y-z)(y+z)3+(z-兀)(z+x)3=-2(兀+y+z)(x一y)(y-z)(z一x)9.在整数集内分解因式(1)—x"—2lx4-45(2)2兀°+7,-2对-13x+6(3)/(y+z)?+y(z+x)2+z(x+y)2-4xyz(4)(x2+1lx+24)(x2+14x+24)-4x2.解：(1)原式=兀彳一兀‘一6兀一15兀=45=x(x2一x-6)-15(%一-=x(x—3)(兀—2)—15(x—3)=(x-3)fx(x-2)-15]=(x-3)(x2+2x-15)=(x-3)(x+5)(x-3)=(x-3)2(x+5)(2)原式=2x4+7x3-4x2+
6、2x2-—13兀+6=%2(2%2+7x—4)+(2x—l)(x—6)=x2(2兀-l)(x+4)+(2x-1)(兀-6)=(2x-1)[x2(x+4)+(x-6)]=(2x・1)(/+4x2+x-6)=(2x・1)(x3+4x2一5兀+6兀一6)=(2x-l)[x(x2+4x—5)+6(2x-1)]=(2x-l)[x(x-l)(x+5)+6(x-1)]=(2x-l)(x-l)[x2+5x4-6]=(2x-l)(x-l)(x+2)(x+3)(3)此多项式是伦换多次试，当x=-y时。原式=0所以有因式(x+y)(y+z)(z+x)设x(y+z)2+y(^+x)2+z(x+y)2-4x)'z=
7、R(x+y)(y+z)(z+x)令x=l,y=2,z=0,得：1.(2+OF+2(0+1)?+0(1+2)2=k.3.2A即6k=6,所以k=lr.x(y+z)2+y(z+x)2+z(x+y)2-4xyz=(x+y)(y+z)(z+x)(4)丿泉式=[(x2+24+1Lr)][C『+24)+14]—4x2=(x2+24)2+14x.(x2+24)+11兀•(兀2+24)+154x2-4x2=(x2+24)2+(x2+24
