證明：

令 a=(1+sqrt(5)/2), b=(1-sqrt(5)/2)

F(n)=[a^n - b^n]/sqrt(5)

F(n+1)^3 + F(n)^3 - F(n-1)^3 = 

[(a^(n+1)-b^(n+1))^3 + (a^(n)-b^(n))^3 + a^(n-1)-b^(n-1))^3]/(5sqrt(5))=

[(a^(3n+3)-3a^(2n+2)b^(n+1)+3a^(n+1)b^(2n+2)-b^(3n+3)) + 

 (a^(3n)-3a^(2n)b^(n)+3a^(n)b^(2n)-b^(3n)) -

 (a^(3n-3)-3a^(2n-2)b^(n-1)+3a^(n-1)b^(2n-2)-b^(3n-3))]/(5sqrt(5))=

[a^(3n-3)(a^6+a^3-1) - 3a^(2n-2)b^(n-1)(a^4b^2+a^2b-1) +

 3a^(n-1)b^(2n-2)(a^2b^4+ab^2-1) - b^(3n-3)(b^6+b^3-1)]/(5sqrt(5))     (1)

可以簡單推出

(a^4b^2+a^2b-1) = (a^2b^4+ab^2-1)=0

(a^6+a^3-1) = 5a^3

(b^6+b^3-1) = 5b^3

代入(1)得

F(n+1)^3 + F(n)^3 - F(n-1)^3 = 

[a^(3n-3)(5a^3) - b^(3n-3)(5b^3)]/(5sqrt(5)) = 

[a^(3n)- b^(3n)]/(sqrt(5)) = F(3n)