Answer: Both sums have the same numerical value, namely \(2015\).

Using the formulae \(\sum_{r=1}^{n}r=\frac{n(n+1)}2\) and \(\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6\),

\(\sum_{r=1}^{13}(3r^2-5r+1)=3\sum_{r=1}^{13}r^2-5\sum_{r=1}^{13}r+13\).

So

\(\sum_{r=1}^{13}(3r^2-5r+1)=3\cdot\frac{13\cdot14\cdot27}{6}-5\cdot\frac{13\cdot14}{2}+13=2015\).

Also, since the term for \(r=0\) is \(-1\),

\(\sum_{r=0}^{9}(r^3-1)=\sum_{r=1}^{9}r^3-10\).

Using \(\sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}2\right)^2\),

\(\sum_{r=0}^{9}(r^3-1)=\left(\frac{9\cdot10}{2}\right)^2-10=45^2-10=2015\).

Therefore the two sums have the same numerical value.