$\\$
`b^2=ac`
`->b.b=a.c`
`->b/a=c/b` (1)
`c^2=bd`
`->c.c=b.d`
`->c/b=d/c` (2)
(1)(2)
`->b/a=c/b=d/c`
Đặt `b/a=c/b=d/c=k(k\ne 0)`
`->b=ak, c=bk,d=ck`
$\\$
`a,`
`(a^3+b^3-c^3)/(b^3+c^3-d^3)`
`= (a^3+b^3-c^3)/(a^3k^3 +b^3k^3 - c^3k^3)`
`=(a^3 +b^3-c^3)/(k^3(a^3+b^3-c^3))`
`= 1/k^3`
`((a+b-c)/(b+c-d))^3`
`=((a+b-c)/(ak+bk-ck))^3`
`=((a+b-c)/(k(a+b-c)))^3`
`= 1/k^3`
Do đó : `(a^3+b^3-c^3)/(b^3+c^3-d^3) =((a+b-c)/(b+c-d))^3(=1/k^3)`
$\\$
`b,`
`a/d`
`=(a.b.c)/(b.c.d)`
`=(a.b.c)/(ak.bk.ck)`
`=(a.b.c)/(a.b.c.k^3)`
`=1/k^3`
`(a^3 + 8b^3 +125c^3)/(b^3+8c^3+125d^3)`
`=(a^3+8b^3+125c^3)/(a^3k^3 + 8b^3k^3 + 125c^3k^3)`
`=(a^3+8b^3+125c^3)/(k^3(a^3+8b^3+125c^3))`
`=1/k^3`
Do đó : `a/d=(a^3 + 8b^3 +125c^3)/(b^3+8c^3+125d^3)(=1/k^3)`
