How to prove that the tensor product of C*-algebras $A \otimes_{\alpha} B$ is simple iff $\alpha$ is the spatial norm and both $A$ and $B$ are simple? This is Exercise 3.4.2 with a hint in Brown and Ozawa's book C star-algebras and finite-dimensional approximations.

Exercise 3.4.2. Prove that $A \otimes_\alpha B$ is simple if and only if
$\|\cdot\|_\alpha=\|\cdot\|_{\min }$ and both $A$ and $B$ are simple.
(Hint: For the "if" direction it suffices to show that every irreducible representation of $A \otimes B$ is faithful. But if $\pi: A \otimes B \rightarrow \mathbb{B}(\mathcal{H})$ is irreducible, then both $\pi_A(A)^{\prime \prime}$ and $\pi_B(B)^{\prime \prime}$ must be factors. One then uses the previous exercise and a theorem of Murray and von Neumann which asserts that if $M \subset \mathbb{B}(\mathcal{H})$ is a factor, then the product $\operatorname{map}$ $$ M \odot M^{\prime} \rightarrow \mathbb{B}(\mathcal{H}) $$ is injective.)

The previous exercise is quoted as follows:

Exercise 3.4.1. Let $\pi: A \otimes B \rightarrow C$ be a $*$-homomorphism which is injective when restricted to $A \odot B$.
Show that $\pi$ must be injective on all of $A \otimes B$. Is this
still true if one replaces $\|\cdot\|_{\min }$ by $\|\cdot\|_{\max }$?

I have worked out the Exercise 3.4.1 and the "only if" part of Exercise 3.4.2, but I cannot get the idea of the hint to do the "if" part. I wonder if someone could give some more clues or elaborate on how we can get the answer from the hint?
Any help is appreciated.
 A: If $J\subset A\otimes B$ is a proper ideal, then $(A\otimes B)/J$ is  a non-zero C$^*$-algebra. Doing GNS for a pure state gives us an irreducible representation $\rho:(A\otimes B)/J\to B(K)$. If we define $\tilde\rho:A\otimes B\to B(K)$ by $\tilde\rho=\rho\circ q$ (where $q$ is the quotient map) we get an irreducible representation of $A\otimes B$ with $J\subset\ker\tilde\rho$. So if we show that any irreducible representation of $A\otimes B$ is faithful, then $A\otimes B$ is simple.
So we assume $\pi:A\otimes B\to B(H)$, irreducible. We have
$$
B(H)=\pi(A\otimes B)''=(\pi_A(A)\cup\pi_B(B))''=(\pi_A(A)'\cap\pi_B(B)')'. 
$$
Taking commutant we get that
$$
\pi_A(A)'\cap\pi_B(B)'=\mathbb C. 
$$
We also know that $\pi_A(A)$ and $\pi_B(B)$ commute, so
$$
\pi_A(A)''\subset \pi_B(B)'. 
$$
It follows that
$$
\pi_A(A)'\cap \pi_A(A)''\subset \pi_A(A)'\cap\pi_B(B)'=\mathbb C. 
$$
This shows that $\pi_A(A)''$ is a factor, and similarly for $\pi_B(B)''$.
Now, using the Murray-von Neumann result, we factor $\pi$ via the  injective maps
$$
A\odot B\xrightarrow{} \pi_A(A)''\odot \pi_A(A)'\xrightarrow{} B(H). 
$$
And then Exercise 3.4.1 finishes the proof.
