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?
 A: This is NOT a one line solution you are expecting but it is an idea which is very useful in solving such determinants which may remind one of Vandermonde matrix.
I'm replacing $a$ by $x$ (for the sake of clarity). Using the linearity of determinants we get 
$$\det=
\begin{vmatrix}
x & x^2 & 1+x^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{vmatrix}
=\begin{vmatrix}
x & x^2 & 1\\
b & b^2 & 1\\
c & c^2 & 1\\
\end{vmatrix}+\begin{vmatrix}
x & x^2 & x^3\\
b & b^2 & b^3\\
c & c^2 & c^3\\
\end{vmatrix}
=A(x)+B(x).
$$
The determinant on the left can be thought of as a third degree polynomial in $x$. Let us call the first determinant (on right) as $A(x)$ (a second degree polynomial in $x$) and the second determinant (on right) as $B(x)$ (a third degree polynomial in $x$).
First consider $A(x)$. Observe that for $x=b$ or $x=c$, this determinant is $0$. Thus both $x-b$ and $x-c$ are factors of this polynomial. Thus 
$$A(x)=K(x-b)(x-c).$$
Moreover 
$$A(0)=bc(c-b) = K(bc).$$
Thus $K=c-b.$ This means
$$A(x)=(c-b)(x-b)(x-c).$$
Likewise
$$B(x)=bc(c-b)x(x-b)(x-c).$$
Thus the given determinant is
$$\det=(c-b)(x-b)(x-c)[1+bcx].$$
Putting back everything in terms of $a$, we get
$$\det=(c-b)(a-b)(a-c)[1+abc].$$
Now you get all the conditions when this can be $0$, namely
$$a=b \quad \text{ or } \quad b=c \quad \text{ or } \quad c=a \quad \text{ or } \quad abc=-1.$$
A: 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:
$$\det (A) = 0 \iff (b = a) \lor (c = a) \lor (c = b) \lor (abc = 1) $$
A: Let $V$ be the Vandermonde matrix in $a,b,c\;$ with the top row being $(1 \; a \; a^2).$ Your determinant is $E=\det(V'')+a b c \det (V)$ where $V''$ is obtained from $V$ by interchanging the first column of $V$ with the second, giving $V',$ and interchanging the second and third columns of $V',$ giving $V''.$ Therefore $\det (V'')=-\det (V')=-(-\det (V)=\det (V)$ and $E=(1+a b c)\det (V)=(1+a b c)(a-b)(b-c)(a-c).$
