You have to determine whether there exist integers

A function is injective if f(a) = f(b) => a = b.

Suppose that f(a) = f(b). Then:

3a³ - a = 3b³ - b

b - a = 3(b - a)(b² + ab + a²)

So (b - a)[3(b² + ab + a²) - 1] = 0.

Here we have a = b as a solution, which would verify that f is injective over the integers. But I'm not convinced because of that quadratic in the other parentheses -- does that matter? Solving 3(b² + ab + a²) - 1 = 0 definitely doesn't lead to a = b...

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