$S^3\times S^5 \simeq \sum X$ for a finite CW-complex $X$ I want to know if $S^3\times S^5$ can be homotopy equivalent to the suspension $\sum X$ of a finite CW-complex $X$. I know the following properties for singular homology (but perhaps it is possible to answer the question without considering homology): 
*for spheres we have $\sum S^n = S^{n+1}$
*$H_k (S^n\times Y) = H_k (Y) \times H_{k-n}(Y)$
*$H_{k+1} (\sum Y) = H_k (Y)$ for $k\geq1$
If we could deduce that the suspension of a finite CW-complex has different homology-modules than $S^3\times S^5$, then we could deduce that they cannot be homotopy equivalent.
Note: If we consider $D^3$ instead of $S^3$ then the claim holds since $D^3$ is contractible (using the first listed property for the suspension of spheres)
 A: As I pointed out in the comments, all cup products in the cohomology ring of a suspension space are trivial. Since there is a non-trivial cup product $H^3(S^3\times S^5)\otimes H^5(S^3\times S^5)\rightarrow H^8(S^3\times S^5)$ given by $(s_3\times 1)\cup(1\times s_5)=s_3\times s_5$, it cannot be that $S^3\times S^5$ has the homotopy type of a suspension.
There is a little more to the story if you are interested. The attaching map for the top cell of any product of spheres $S^m\times S^n$ is given by the Whitehead product $w:S^{m+n-1}\rightarrow S^m\vee S^n$, which you can write down explictly if you want (try first with $m=n=1$). This is a surprisingly fun little map, which is never a suspension, and moreover is stably trivial. In fact a single suspension suffices to kill it.
Below, you'll find a purely homotopy-theoretic disproof of your question, which I'll include for fun.
There are projections $pr_1:S^3\times S^5\rightarrow S^3$, $pr_2:S^3\times S^5\rightarrow S^5$ and $q:S^3\times S^5\rightarrow S^3\wedge S^5\cong S^8$ and algebraically these induce an isomorphism of abelian groups 
$$pr_1\oplus pr_2\oplus q:H_*(S^3\times S^5)\cong H_*S^3\oplus H_*S^5\oplus H_*S^8\cong H_*(S^3\vee S^5\vee S^8).$$
Observe that this isomorphism has been algebraically constructed, and does not come from one of spaces.
However, if we assume that $S^3\times S^5$ is homotopy equivalent to the suspension of some space $X$, then this equivalence transfers the suspension co-multiplication on $\Sigma X$ to a co-H-structure on $S^3\times S^5$. We can then use this to add the the three previous maps to get a map
$$pr_1+pr_2+q:S^3\times S^5\rightarrow S^3\vee S^5\vee S^8$$
which realises the previous abstract isomorphism of integral homology modules topologically. Then, since both the target and domain of this map are simply-connected (check their cellular constructions), the Whitehead theorem tells us that this map is a homotopy equivalence.
But this is absurd, since the inclusion $S^8\hookrightarrow S^3\vee S^5\vee S^8$ generates a free $\mathbb{Z}$ summand in $\pi_8$, whilst $\pi_8(S^3\times S^5)\cong \pi_8S^3\oplus\pi_8S^5\cong\mathbb{Z}_2\oplus\mathbb{Z}_{24}$ is finite. Hence we have a contradiction.
