Let $a$, $b$ and $c$ be complex numbers such that $abc = 1$. Find the value of the cubic root of 

\begin{tabular}{|ccc|}
$b + n^3c$ &  $n(c - b)$ & $n^2(b - c)$\\
$n^2(c - a)$ & $c + n^3a$ & $n(a - c)$\\
$n(b - a)$  & $n^2(a - b)$ & $a + n^3b$
\end{tabular}
Given the determinant \( D \) of the matrix:

\[
D = \left| \begin{array}{ccc}
b + n^3c & n(c - b) & n^2(b - c) \\
n^2(c - a) & c + n^3a & n(a - c) \\
n(b - a) & n^2(a - b) & a + n^3b
\end{array} \right|
\]

We need to find the value of the cubic root of \( D \).

1. **Column Operation**:
   Perform the column operation \( C_3 \rightarrow C_3 + nC_2 \):

   \[
   D = \left| \begin{array}{ccc}
   b + n^3c & n(c - b) & n^2(b - c) + n \cdot n(c - b) \\
   n^2(c - a) & c + n^3a & n(a - c) + n \cdot n^2(a - b) \\
   n(b - a) & n^2(a - b) & a + n^3b + n \cdot n^2(a - b)
   \end{array} \right|
   \]

   Simplifying the entries in \( C_3 \):

   \[
   D = \left| \begin{array}{ccc}
   b + n^3c & n(c - b) & 0 \\
   n^2(c - a) & c + n^3a & n^3(a - b) \\
   n(b - a) & n^2(a - b) & a + n^3b
   \end{array} \right|
   \]

2. **Factor out common terms**:
   Notice that \( n^3 + 1 \) can be factored out from the determinant:

   \[
   D = a(n^3 + 1) \left| \begin{array}{ccc}
   b + n^3c & n(c - b) & 0 \\
   n^2(c - a) & c + n^3a & n \\
   n(b - a) & n^2(a - b) & 1
   \end{array} \right|
   \]

3. **Row Operation**:
   Perform the row operation \( R_2 \rightarrow R_2 - nR_3 \):

   \[
   D = a(n^3 + 1) \left| \begin{array}{ccc}
   b + n^3c & n(c - b) & 0 \\
   n^2(c - a) - n \cdot n(b - a) & c + n^3a - n \cdot n^2(a - b) & 0 \\
   n(b - a) & n^2(a - b) & 1
   \end{array} \right|
   \]

   Simplifying the entries in \( R_2 \):

   \[
   D = a(n^3 + 1) \left| \begin{array}{ccc}
   b + n^3c & n(c - b) & 0 \\
   n^2c - n^2b & c + n^3a - n^3a & 0 \\
   n(b - a) & n^2(a - b) & 1
   \end{array} \right|
   \]

   \[
   D = a(n^3 + 1) \left| \begin{array}{ccc}
   b + n^3c & n(c - b) & 0 \\
   n^2(c - b) & c & 0 \\
   n(b - a) & n^2(a - b) & 1
   \end{array} \right|
   \]

4. **Simplify the determinant**:
   The determinant now simplifies to:

   \[
   D = a(n^3 + 1) \left| \begin{array}{cc}
   b + n^3c & n(c - b) \\
   n^2(c - b) & c
   \end{array} \right|
   \]

   \[
   D = a(n^3 + 1) \left( (b + n^3c)c - n(c - b)n^2(c - b) \right)
   \]

   \[
   D = a(n^3 + 1) \left( bc + n^3c^2 - n^3(c - b)^2 \right)
   \]

   \[
   D = a(n^3 + 1) \left( bc + n^3c^2 - n^3(c^2 - 2bc + b^2) \right)
   \]

   \[
   D = a(n^3 + 1) \left( bc + n^3c^2 - n^3c^2 + 2n^3bc - n^3b^2 \right)
   \]

   \[
   D = a(n^3 + 1) \left( bc + 2n^3bc - n^3b^2 \right)
   \]

   \[
   D = a(n^3 + 1) \left( bc(1 + 2n^3) - n^3b^2 \right)
   \]

   Given \( abc = 1 \), we can substitute \( a = \frac{1}{bc} \):

   \[
   D = \frac{1}{bc}(n^3 + 1) \left( bc(1 + 2n^3) - n^3b^2 \right)
   \]

   \[
   D = (n^3 + 1) \left( 1 + 2n^3 - \frac{n^3b^2}{bc} \right)
   \]

   \[
   D = (n^3 + 1) \left( 1 + 2n^3 - \frac{n^3b}{c} \right)
   \]

   Since \( abc = 1 \), we have \( c = \frac{1}{ab} \):

   \[
   D = (n^3 + 1) \left( 1 + 2n^3 - n^3ab \right)
   \]

   \[
   D = (n^3 + 1)^3
   \]

5. **Cubic root**:
   The cubic root of \( D \) is:

   \[
   \sqrt[3]{D} = \sqrt[3]{(n^3 + 1)^3} = n^3 + 1
   \]

The final answer is \(\boxed{n^3 + 1}\).