It's easy$^*$ to prove that if $n=3^{6m}(3k \pm 1)$ where $(m,k) \in \mathbb{N} \times \mathbb{Z}$, then $n=a^2+b^2+c^3+d^3$ with $(a,b,c,d) \in \mathbb{Z}^4$. But how to prove that this is true if $n=3k$?
Thanks,
W
$^*$ Because $3k+1=0^2+(3k+8)^2+(k+1)^3+(-k-4)^3$, $3k+2=1^2+(3k+8)^2+(k+1)^3+(-k-4)^3$ and $3^6(a^2+b^2+c^3+d^3)=(27a)^2+(27b)^2+(9c)^3+(9d)^3$.