Gọi A=8.(x+y+z)3-(x+y)3-(x+z)3-(y+z)3
Đặt a=x+y
b=x+z
c=y+z
=>\(\dfrac{a+b+c}{2}=x+y+z\)
Do đó :
A=8(\(\dfrac{a+b+c}{2}\))3-a3-b3-c3
=8.\(\dfrac{\left(a+b+c\right)^3}{8}\)-a3-b3-c3
=(a+b+c)3-a3-b3-c3
=(a+b)3+c3+3(a+b)c(a+b+c)-(a+b)3-c3+3ab(a+b)
=3(a+b)(ca+cb+c2+ab)
=3(a+b)[(ca+c2)+(cb+ab)]
=3(a+b)[c(a+c)+b(a+c)]
=3(a+b)(c+b)(b+c)
Thay
a=x+y
b=x+z
c=y+z
vào trên ta được:
A=3(2x+y+z)(2y+x+z)(2z+x+y)
Vậy 8.(x+y+z)3-(x+y)3-(x+z)3-(y+z)3 = 3(2x+y+z)(2y+x+z)(2z+x+y)
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