If $D_p=\left|\begin{array}{ccc}p& 15& 8\\ p^2& 35& 9\\ p^3& 25& 10\end{array}\right|$, then $D_1+D_2+D_3+D_4+D_5=$

Due to the linearity property of determinants, the sum of determinants that differ only in one column is the determinant of the matrix with that column replaced by the sum of corresponding columns.
$$\sum _{p=1}^5{D}_p=\left|\begin{array}{ccc}\sum p& 15& 8\\ \sum p^2& 35& 9\\ \sum p^3& 25& 10\end{array}\right|$$
Calculate the sums for $p=1$ to $5$:$\sum p=\frac{5\times 6}{2}=15$$\sum p^2=\frac{5\times 6\times 11}{6}=55$$\sum p^3={\left(\frac{5\times 6}{2}\right)}^2=15^2=225$ Substitute into the determinant:
$$\left|\begin{array}{ccc}15& 15& 8\\ 55& 35& 9\\ 225& 25& 10\end{array}\right|$$
Take 5 out from column 1 and 5 out from column 2:
$$25\left|\begin{array}{ccc}3& 3& 8\\ 11& 7& 9\\ 45& 5& 10\end{array}\right|$$
Calculating the determinant:$=25[3(70-45)-3(110-405)+8(55-315)]$$=25[3(25)-3(-295)+8(-260)]$$=25[75+885-2080]=25[-1120]=-28000$ None of the options match $-28000$.