<s>[INST] Find the determinant without row expansion Show that the determinant of the matrix 
\begin{bmatrix} 1& a& a^3\\
1& b& b^3\\  
1& c& c^3\end{bmatrix}
is $(a-b)(b-c)(c-a)(a+b+c)$ without expanding.
I was able to get out $(a-b)(b-c)(c-a)$ but couldn't complete.
 [/INST] Use row operations to simplify. In the process, the factorization drops right out.
\begin{align}
\det \begin{bmatrix}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3
\end{bmatrix}
&= \det \begin{bmatrix}
1 & a & a^3 \\
0 & b-a & b^3-a^3 \\
0 & c-a & c^3-a^3
\end{bmatrix} \\
&= \det \begin{bmatrix}
1 & a & a^3 \\
0 & b-a & (b-a)(b^2+ab+b^2) \\
0 & c-a & (c-a)(c^2+ac+a^2)
\end{bmatrix} \\
&= (b-a)(c-a)\det \begin{bmatrix}
1 & a & a^3 \\
0 & 1 & b^2+ab+a^2 \\
0 & 1 & c^2+ac+a^2
\end{bmatrix} \\
&= (b-a)(c-a)\det \begin{bmatrix}
1 & a & a^3 \\
0 & 1 & b^2+ab+a^2 \\
0 & 0 & c^2-b^2+ac-ab
\end{bmatrix} \\
&= (b-a)(c-a)(c^2-b^2 + ac-ab)\det \begin{bmatrix}
1 & a & a^3 \\
0 & 1 & b^2+ab+a^2 \\
0 & 0 & 1
\end{bmatrix} \\
&= (b-a)(c-a)(c-b)(c+b+a) \\
&= (a-b)(b-c)(c-a)(a+b+c).
\end{align}
</s>