46 (1) a について整理すると
(S)
(a+b+c)(a² + b²+c²-ab-bc-ca)
10+ st= = {a+(b+c)}{a² − (b+c)a+ (b² − bc+c²)}
= a³ − (b+c)a² + (b² − bc+c²)a+(b+c)a² – (b+c)³a
= a³ +
= a³ + (-3bc)a+b³ + c³
TRA
{(b²-bc+c²2)-(b²+2bc+c²)}a+(b³c³)
22 +(b+c)(b²-bc+c²)
のどこ?
- dat
gees. [= ês.1 -
(²_bc+c²) a
S
_C0 = a³ + b³ + c³-3abc
(2) (1)より
a³ + b³ + c³-3abc = (a+b+c)(a² +6²+c²-ab-bc-ca)
a,b,c をx,y, 2zにおきかえると
LO
x+y+82-6xyz=(x+y+2z)(x^2+y2+4z²-xv-2v²-2zr)
al-(b + c)² a
= {(6²-bc+c²)
XE643-6²2bc+c²)}a
(b+c)(b²-bc+c²)
=b³ + c³ EI
23