Proof:
Let G be a group and f: G → G be an automorphism defined by f(x) = x³.

1. Claim: G is abelian.
Proof:
To prove that G is abelian, we need to show that for any two elements a and b in G, ab = ba.

Let a, b ∈ G be arbitrary. Since f is an automorphism, it is bijective, which means it is both injective and surjective.

2. Claim: f is injective.
Proof:
Suppose f(a) = f(b) for some a, b ∈ G. Then, by the definition of f, we have a³ = b³.

Taking the cube root of both sides, we get a = b. Thus, f is injective.

3. Claim: f is surjective.
Proof:
Let y ∈ G be arbitrary. We need to show that there exists an element x ∈ G such that f(x) = y.

Since f is a mapping from G to G, there exists an element x = ∛y such that f(x) = (∛y)³ = y. Thus, f is surjective.

4. Claim: f is a homomorphism.
Proof:
To prove that f is a homomorphism, we need to show that for any two elements a and b in G, f(ab) = f(a)f(b).

Let a, b ∈ G be arbitrary. We have f(ab) = (ab)³ = a³b³.

Since G is a group, it satisfies the associative property, so we can rewrite this as f(ab) = a³b³ = (a³)(b³) = f(a)f(b).

Therefore, f is a homomorphism.

5. Claim: f is a monomorphism.
Proof:
Since f is injective and a homomorphism, it is also a monomorphism.

Conclusion:
Since f is a monomorphism, it preserves the group structure. Therefore, G must be isomorphic to the group G' = {x³ | x ∈ G} under the operation of multiplication.

In G', the operation of multiplication is commutative because for any two elements a³ and b³ in G', we have (a³)(b³) = a³b³ = b³a³ = (b³)(a³).

Therefore, G' is an abelian group. Since G is isomorphic to G', G must also be abelian.

Hence, we have proved that if f: G → G given by f(x) = x³ is an automorphism, then G is abelian.