4. Find the sum: $S_{n}=2^{3}+5^{3}+8^{3}+\cdots+(3 n-1)^{3}$.
4. The difference equation is $f(x+1)-f(x)=(3 x+2)^{3}$, and $f(1)=8$. Let the particular solution be $R(x)=x \cdot Q_{3}(x)=A x^{4}+$ $B x^{3}+C x^{2}+D x$. Substituting into the difference equation and simplifying, we get $4 A x^{3}+(6 A+3 B) x^{2}+(4 A+3 B+2 C) x+(A+B+C+D)=$ $27 x^{3}+54 x^{2}+36 x+8$. By comparing the coefficients of like terms on both sides, we obtain: $4 A=27,6 A+3 B=54,4 A=3 B+2 C=36, A+$ $B+C+D=8$. Solving these, we get $A=\frac{27}{4}, B=\frac{9}{2}, C=-\frac{9}{4}, D=-1$. In the general solution $f(x)=C \cdot 1^{x}+R(x)$, setting $x=1$, we get $8=c+8$, hence $c=0$. Therefore, the summation formula is $S_{n}=\frac{n}{4}\left(27 n^{3}+18 n^{2}-9 n-4\right)$.