Answer
$abc-a^3-b^3-c^3$
Work Step by Step
We know that for a matrix
\[
\left[\begin{array}{rrr}
a & b & c \\
d &e & f \\
g &h & i \\
\end{array} \right]
\]
the determinant, $D=a(ei-fh)-b(di-fg)+c(dh-eg).$
Hence here $D=a(cb-a^2)-b(b^2-ac)+c(ba-c^2)=abc-a^3-b^3-abc+abc-c^3=abc-a^3-b^3-c^3$
