I'm working my way through Gallian's Contemporary Abstract Algebra (9th edition). In chapter six which is about isomorphisms there is the following example on page 123
Example 3: The mapping from $\mathbb{R}$ under addition to itself given by $\phi(x) = x^3$ is not an isomorphism. Although $\phi$ is one-to-one and onto it is not operation preserving, since it is not true that $(x + y)^3 = x^3 + y^3$ for all $x, y$.
My problem is with the operation preserving portion. Here is my solution:
Solution: Write $G = (\mathbb{R}, +)$ and $\bar{G} = (\mathbb{R}, +)$. Let $\phi : G \to \bar{G}$ such that $x \mapsto x^3$. To check whether $\phi$ is operation preserving, we must determine whether $\phi(x + y) = \phi(x) + \phi(y)$ holds for all $x, y \in G$. We get
\begin{aligned} \phi(x + y) &= (x + y)^3 \\ &= 3(x + y) & \mbox{(additive)} \\ &= (x + y) + (x + y) + (x + y) \\ &= 3x + 3y & \mbox{(addition is commutative)} \\ &= x^3 + y^3 & \mbox{(additive)} \\ &= \phi(x) + \phi(y) \end{aligned}
Which shows that $\phi(x)$ is an isomorphism contrary to Gallian. This is for self-study so I would appreciate a hint.
Gallian's intent, as pointed out by @AymanHourieh in the first comment, is that $x\mapsto x^3$ is understood to be multiplicative, not additive.
Then $$\begin{align}\phi(x+y)&=(x+y)^3 \\ &=x^3+3x^2y+3xy^2+y^3 \\ &\neq x^3+y^3 \tag{1} \\ &=\phi(x)+\phi(y),\end{align}$$ where $(1)$ holds when, say, $x=y=42$.
Hence $\phi$ is not an isomorphism.