How to compute $K(X\times S^2)$ using $\Omega ^2 BU\simeq BU\times \mathbb{Z}$ I'm reading "Algebraic Topology from a Homotopical Viewpoint."
The exercise 9.5.14 of it is as follows:
Deduce from the Bott periodicity theorem (in this book it means $\Omega ^2 BU\simeq BU\times \mathbb{Z}$) that $K(X\times S^2)$ is isomophic to the free module over the ring $K(X)$ with two generators $1$ and $L-1$.
Here $L$ denotes the pullback of the canonical line bundle over $S^2\cong CP^1$.
It seems that this is a standard approach for proving the Bott periodicity, but in this book one proves it by constructing the quasifibration $BU\times \mathbb{Z}\to E\to U$ where E is a contractible space.
I think it's not very difficult to prove that $K(X\times S^2)\cong K(X)\oplus K(X)$ using $\tilde{K}(X\times Y)\cong \tilde{K}(X)\oplus \tilde{K}(Y)\oplus \tilde{K}(X\wedge Y)$, but I have no idea how to find the generator.
What might be the answer authors intended?
 A: Here's at least what I imagine how the authors wanted the solution to start.  Recall that $K$-theory is "represented", i.e., $$K(X) \cong [X_+, BU \times \mathbb{Z}].$$  So,
\begin{align*}
K(X \times S^2) &\cong [(X \times S^2)_+, BU \times \mathbb{Z}] \\ &\cong [(X_+ \wedge S^2) \vee X_+, BU \times \mathbb{Z}] \\ &\cong [X_+ \wedge S^2, BU \times \mathbb{Z}] \oplus [X_+, BU \times \mathbb{Z}].
\end{align*}
The second summand is a copy of $K(X)$, generated by $1$, as you can check using the projection $X \times S^2 \to X$.  The first summand is trickier.  By adjunction and then the form of Bott periodicity you stated, we have $$[X_+ \wedge S^2, BU \times \mathbb{Z}] \cong [X_+, \Omega^2(BU \times \mathbb{Z})] \cong [X_+, BU \times \mathbb{Z}] \cong K(X).$$
To identify the generator, you need to know something about the map that implements the Bott isomorphism $BU \times \mathbb{Z} \xrightarrow{\sim} \Omega^2 BU$.  For example, you might know that it's represented by tensoring with $L$ (or perhaps $L - 1$).  You might get away with just knowing $L$ generates $K(S^2)$ or $\pi_2 BU$.  
Once you know this, the conclusion of the problem follows.  
A: $(X \times S^2)_{+}$ is in general not homotopy equivalent to
$(X_{+} \wedge S^2) \vee X_{+}$.
Just take $X:= \{x_{1}, x_{2} \}$:
Then $(X \times S^2)_{+}$ is the disjoint union of two copies of $S^2$ with a disjoint basepoint.
And $(X_{+} \wedge S^2) \vee X_{+}$ is $S^2 \vee S^2$ and two disjoint points.
These two spaces are obviously not homotopy equivalent.
