Calculating the determinant as a product without making any calculations My problem is on the specific determinant.
$$\det
\begin{pmatrix}
  na_1+b_1 & na_2+b_2 & na_3+b_3 \\
  nb_1+c_1 & nb_2+c_2 & nb_3+c_3 \\
  nc_1+a_1 & nc_2+a_2 & nc_3+a_3 
\end{pmatrix}=
(n+1)(n^2-n+1)
\det\begin{pmatrix}
  a_1 & a_2 & a_3 \\
  b_1 & b_2 & b_3 \\
  c_1 & c_2 & c_3  
\end{pmatrix}$$
All I can do is prove the factor $(n+1)$ and I think that we have to work only on one column and the do the exact same thing to the others. 
 A: Use multilinearity of determinant:
$$\begin{vmatrix}na_1+b_1&na_2+b_2&na_3+b_3\\
nb_1+c_1&nb_2+c_2&nb_3+c_3\\
nc_1+a_1&nc_2+a_2&nc_3+a_3\end{vmatrix}=\begin{vmatrix}na_1&na_2&na_3\\
nb_1&nb_2&nb_3\\
nc_1&nc_2&nc_3\end{vmatrix}+\begin{vmatrix}na_1&na_2&b_3\\
nb_1&nb_2&c_3\\
nc_1&nc_2&a_3\end{vmatrix}+$$
$$+\begin{vmatrix}na_1&b_2&na_3\\
nb_1&c_2&nb_3\\
nc_1&a_2&nc_3\end{vmatrix}+\begin{vmatrix}na_1&b_2&b_3\\
nb_1&c_2&c_3\\
nc_1&a_2&a_3\end{vmatrix}+\begin{vmatrix}b_1&na_2&na_3\\
c_1&nb_2&nb_3\\
a_1&nc_2&nc_3\end{vmatrix}+\ldots$$
Observe that if we put
$$\Delta=\begin{vmatrix}a_1&a_2&a_3\\
b_1&b_2&b_3\\
c_1&c_2&c_3\end{vmatrix}$$
then we have that the four first determinants above equal (factor out constants from rows/columns):
$$n^3\Delta+n^2\Delta+n^2\begin{vmatrix}a_1&b_2&a_3\\b_1&c_2&b_3\\c_1&a_2&c_3\end{vmatrix}+n\begin{vmatrix}a_1&b_2&b_3\\b_1&c_2&c_3\\c_1&a_2&a_3\end{vmatrix}+\ldots$$
Well, develop the other three determinants left and sum up all.
A: If you notice that 
$$
\begin{pmatrix}
  na_1+b_1 & na_2+b_2 & na_3+b_3 \\
  nb_1+c_1 & nb_2+c_2 & nb_3+c_3 \\
  nc_1+a_1 & nc_2+a_2 & nc_3+a_3
\end{pmatrix}=
\begin{pmatrix}
  n & 1 & 0 \\
  0 & n & 1 \\
  1 & 0 & n \\
\end{pmatrix}
\begin{pmatrix}
  a_1 & a_2 & a_3 \\
  b_1 & b_2 & b_3 \\
  c_1 & c_2 & c_3
\end{pmatrix}$$
and if you know that $\det(AB)=\det(A)\det(B)$, then it only remains to calculate determinant of the matrix
$\begin{pmatrix}
  n & 1 & 0 \\
  0 & n & 1 \\
  1 & 0 & n \\
\end{pmatrix}$.
It is not very difficult to notice the above identity, if you know, that multiplication (from the left) means simply making linear combinations of the rows. See, for example, this answer.
