if f(x) is a one-to-one function, then g(x) = f(x^3) is also a one-to-one function
The Attempt at a Solution
Assume f(x) is a one to one function. For a particular element of B, there is at most one element of A which is mapped to it. Therefore, by the definition of injective x^3 must also be one to one.
This is what I have, but I don't think that this is the way to go about it.