Answer
$$-123165$$
Work Step by Step
Given $$\sum_{m=1}^{30}(4-m)^{3}$$
Since
\begin{align}
\sum_{j=1}^{N} j&=1+2+\cdots+N=\frac{N(N+1)}{2} \tag{1}\\
\sum_{j=1}^{N} j^{2}&=1^{2}+2^{2}+\cdots+N^{2}=\frac{N(N+1)(2 N+1)}{6} \tag{2}\\
\sum_{j=1}^{N} j^{3}&=1^{3}+2^{3}+\cdots+N^{3}=\frac{N^{2}(N+1)^{2}}{4} \tag{3}
\end{align}
Then
\begin{align*}
\sum_{m=1}^{30}(4-m)^{3}&= \sum_{m=1}^{30}\left(64-m^{3}-48 m+12 m^{2}\right)\\
&=\sum_{m=1}^{30} 64-\sum_{m=1}^{30} m^{3}-\sum_{m=1}^{30} 48 m+\sum_{m=1}^{30} 12 m^{2}\\
&=\sum_{m=1}^{30} 64-\sum_{m=1}^{30} m^{3}-48 \sum_{m=1}^{30} m+12 \sum_{m=1}^{30} m^{2}, \ \text{Use (1), (2), (3)}\\
&= (64)(30)-\frac{30^{2}(30+1)^{2}}{4} -48\frac{30(30+1)}{2}+12 \frac{30(30+1)(2 (30)+1)}{6} \\
&= -123165
\end{align*}
