One of the following equalities is true.
(a) $3 \cdot \omega = \omega \cdot 3$
(b) $\omega \cdot 2 + \omega^2 = \omega \cdot 3 + 2 + \omega^2$
(c) $\omega^2 + \omega^3 = \omega^2 + \omega^2$
(d) $4 \cdot (3 + \omega) = (3 + \omega) \cdot 4$
I think the answer is (c), but I am not sure. So (a) is wrong because $3 \cdot \omega$ is $\omega$ and $\omega \cdot 3 = \omega \cdot 2 + \omega = \omega + \omega + \omega$. (d) is wrong for a similar reason when we simplify $(3 + \omega) = \omega$. For (b) $\omega \cdot 3 + 2 = \omega + \omega + (\omega + 2) = \omega + \omega + (SS \omega) \neq \omega + \omega$.
For (c), it does not look correct to me because $\omega^2 = \omega \cdot \omega$ and $\omega^3 = \omega \cdot \omega \cdot \omega$. However, that means no answer is correct. Did I misunderstand something in my derivation?
$a)$ is false since $$3\cdot \omega=\sup\{3n:n\in\mathbb{N}\}=\omega,$$ and $$\omega\cdot 3=\omega(1+1+1)=\omega+\omega+\omega>\omega.$$ $c)$ is false since $$\omega^2+\omega^3=\omega^2(1+\omega)=\omega^2\cdot\omega=\omega^3,$$ and $$\omega^2+\omega^2=\omega^2\cdot 2<\omega^3.$$ $d)$ is false since $$4\cdot (3+\omega)=4\cdot\sup\{3+n:n\in\mathbb{N}\}=4\cdot \omega=\sup\{4n:n\in\mathbb{N}\}=\omega,$$ and $$(3+\omega)\cdot 4=\omega\cdot 4=\omega+\omega+\omega+\omega>\omega.$$ $b)$ is true since $$\omega\cdot 2+\omega^2=\omega(2+\omega)=\omega\cdot\omega=\omega^2,$$ and $$\omega\cdot 3+2+\omega^2=S(S(\omega+\omega+\omega))+\omega^2=\omega^2.$$