= Solution
For a <continuous map> $f:S^2\times S^4\to\mathbb{CP}^3$, use the classes from part (a). Since $H^2(S^2\times S^4;\mathbb Z)=\mathbb Zu$, there is an <integer> $k$ with $f^*c=ku$. Naturality of the <cup product> and $u^2=0$ yield
$$
f^*(c^3)=(f^*c)^3=k^3u^3=0.
$$
With the product <orientation> on the source and the complex <orientation> on the target, the <mapping degree> $d$ is defined by $f_*[S^2\times S^4]=d[\mathbb{CP}^3]$. Evaluation on the <fundamental class> gives
$$
d=\langle c^3,f_*[S^2\times S^4]\rangle
=\langle f^*(c^3),[S^2\times S^4]\rangle=0.
$$
Thus \b[the induced <group homomorphism> on $H_6$ is zero]. This is a <cup-power obstruction to nonzero degree>, rather than a consequence of the additive <homology groups> alone.
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