a1(4) 00解
1b 1001c 100.1d
0r1+ar201+ab0===== 1b
0 11
d00
a
1001b 1001c 1a1c 1001d
+aba0c3+dc2+abaad
=( 1)( 1)2+1 1c1===== 1c1+cd
0 1d0 10
abad=abcd+ab+cd+ad+1.=( 1)( 1)3+2+ 11+cd
5.证明:
a2abb2
(1)2aa+b2b=(a b)3;
111
证明
a2abb2c2 c1a2ab a2b2 a22aa+b2b=====2ab a2b 2a
00111c3 c11
222
ab ab ab+a=(a b)3.=(b a)(b a)a=( 1)12b a2b 2a
3+1
ax+byay+bzaz+bxxyz
(2)ay+bzaz+bxax+by=(a3+b3)yzx;az+bxax+byay+bzzxy
证明
ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz