数学 展開

質問

(a+b+c)^3=a^3+b^3+c^3+3a^2b+3b^2c+3c^2a+3ab^2+3bc^2+3ca^2+6abc

(a-b-c)^3=a^3-b^3-c^3-3a^2b-3b^2c+3c^2a+3ab^2-3bc^2-3ca^2+6abc

ですか？

回答No.2

bran111

b+c=dとおくと

(a+b+c)^3=(a+d)^3=a^3+3a^2d+3ad^2+d^3
=a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^3
=a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+c^3+3b^2c+3bc^2
=a^3+b^3+c^3+3a^2b+3b^2c+3c^2a+3ab^2+3bc^2+3ca^2+6abc

(a-b-c)^3=(a-d)^3=a^3-3a^2d+3ad^2-d^3
=a^3-3a^2(b+c)+3a(b+c)^2-(b+c)^3
=a^3-3a^2b-3a^2c+3ab^2+6abc+3ac^2-b^3-c^3-3b^2c-3bc^2
=a^3-b^3-c^3-3a^2b-3b^2c+3c^2a+3ab^2-3bc^2-3ca^2+6abc

回答No.1

info222_

２式とも、あっています。