<s>[INST] To obtain the condition for vanishing of the given determinant If $a,b,c$ are distinct real numbers obtain the condition under which the following determinant vanishes. 
$$\left|
\begin{array}{cc} 
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{array}
\right|$$
My answer: After a little calculation I was able to show that $D=0$ reduces to $abc=-1$. 
Is there a simpler one line answer to this? especially since this matrix looks suspiciously similar to the Vandermonde matrix?
 [/INST] There is a pretty easy way to obtain the determinant:
First, use the linearity of the determinant to split up the matrix: $$\det (A)=
\begin{vmatrix}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{vmatrix}
=\begin{vmatrix}
a & a^2 & 1\\
b & b^2 & 1\\
c & c^2 & 1\\
\end{vmatrix}+\begin{vmatrix}
a & a^2 & a^3\\
b & b^2 & b^3\\
c & c^2 & c^3\\
\end{vmatrix}$$
Then, use $\det(A \cdot B) = \det(A) \cdot \det(B)$ to split up the right matrix; also, switch the colums in the left matrix. What you get are two identical Vandermonde matrices and a diagonal matrix:
$$\det (A)=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}+\begin{vmatrix}
a & 0 & 0\\
0 & b & 0\\
0 & 0 & c\\
\end{vmatrix}\begin{vmatrix}
1 & a & a^2\\
1 & b & b^2\\
1 & c & c^2\\
\end{vmatrix}$$
Now, you can get the determinants with the Vandermonde formula and the formula for diagonal matrices:
$$\det (A)=(b-a)(c-a)(c-b)+abc((b-a)(c-a)(c-b))$$
$$=(b-a)(c-a)(c-b)(abc+1)$$
Now you can easily see when $\det (A)$ vanishes:
$$\det (A) = 0 \iff (b = a) \lor (c = a) \lor (c = b) \lor (abc = 1) $$
</s>