I was trying to show that $V^{\otimes 2}=S^2(V)\oplus \Lambda^2V$ and this is what I came up with. Obviously $S^2(V)\oplus \Lambda^2V\subset V^{\otimes 2}$. Now take $a\in V^{\otimes 2}$.Let $\dim V=n$ and let the basis of $V^{\otimes 2}$ be $\{v_i\otimes v_j\}_{1\leq i,j \leq n}$. Then we can write $a=\sum\frac{1}{2}a_{ij}((v_i\otimes v_j)+(v_j\otimes v_i))+\sum\frac{1}{2}a_{ij}((v_i\otimes v_j)-(v_j\otimes v_i))$ and the first one is in $S^2(V)$ and the other in $\Lambda^2V$. Also, we see that intersection of $S^2(V)$ and $\Lambda^2V$ is just $\{0\}$.
I hope what I did was right. My question is: Can we say that $V^{\otimes k}=S^k (V)\oplus \Lambda^k V$ for say $k=3$ or higher? My intuition says no because I can't write any element in $V^{\otimes k}$ into two pieces like I did before.