If `a , b` are two fixed positive integers such that `f(a+x)=b+[b^3+1-3b^2f(x)+3b{f(x)}^2-{f(x)}^3]^(1/3)` for all real `x ,` then prove that `f(x)` is periodic and find its period.

Explanation

`f(a+x)=b+[b^(3)+1-3b^(2)f(x)+3b{f(x)}^(2)-{f(x)}^(3)]^(1//3)`
`=b+[1+{b-f(x)}^(3)]^(1//3)`
`or f(a+x)-b=[1-{b-f(x)}^(3)]^(1//3)`
` or phi(a+x)=[1-{phi(x)}^(3)]^(1//3) " (1)" `
where `phi (x)=f(x)-b`
`or phi (2a+x)=[1-{phi(x+a)}^(3)]^(1//3)=phi(x) " From (1)]" `
`or f(x+2a)-b=f(x)-b`
`or f(x+2a)=f(x)`
Therefore, `f(x)` is periodic with period 2a.