Question

已知a+b+c+d=0,求证:a^3+b^3+c^3+d^3=3(a+b)(b+c)(c+d)

Answer 1

a^3+b^3=(a+b)*(a^2-ab+b^2)
a^3+b^3+c^3+d^3
=(a+b)*(a^2-ab+b^2)+(c+d)(c^2-cd+d^2)
=(a+b)(a^2-ab+b^2-c^2+cd-d^2) [c+d=-(a+b)]
=(a+b)[(b+c)(b-c)+(a+d)(a-d)-ab-bc+bc+cd]
=(a+b)[(b+c)(b-c-a+d)-b(a+c)+c(b+d)] [a+d=-(b+c)]
=(a+b)[(b+c)(b-c-a+d)+(b+d)(b+c)] [a+c=-(b+d)]
=(a+b)(b+c)(b-c-a+d+b+d)
=3(a+b)(b+c)(b+d)

Answer 2

题错！
例如：
a=2
b=1
c=-2
d=-1
原式左边=8+1-8-1=0
原式右边=3×3×(-1)×(-3)