Question

已知x/a=y/b=z/c,且a+b+c≠0,求证(x/a)^3+(y/b)^3+(z/c)^3=3[(x+y+z)/(a+b+c)]^3

Answer 1

简单
x/a=y/b=z/c,a+b+c≠0 (x/a)^3+(y/b)^3+(z/c)^3=3(x/a)^3
3(x/a)^3-3[(x+y+z)/(a+b+c)]^3=3[x/a-(x+y+z)/(a+b+c)]^3
x/a-(x+y+z)/(a+b+c)=[a(x+y+z)-(a+b+c)x]/a(a+b+c)
a(x+y+z)-(a+b+c)x=ay+az-bx-cx
由于x/a=y/b=z/c
所以xb=ay xc=az yc=zb
所以ay+az-bx-cx=0
即(x/a)^3+(y/b)^3+(z/c)^3-3[(x+y+z)/(a+b+c)]^3=0
即问题被证明了