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?

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
$$a=b \quad \text{ or } \quad b=c \quad \text{ or } \quad c=a \quad \text{ or } \quad abc=-1.$$
