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.
 A: 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}
A: Hint: reduce the matrix to upper triangular form and then read off the determinant as the product of the diagonals.
A: \begin{bmatrix} 1& a& a^2&a^3\\
1& b& b^2 &b^3\\  
1& c& c^2 &c^3\\
1& X& X^2 &X^3
\end{bmatrix}is the well known Vandermonde determinant. When expend with respect to the last line, this is a polynomial
whose $X^2$ coefficient is the opposite of the result.
Now, expending 
$$
(c-a)(b-a)(c-b)(X-c)(X-b)(X-a) \\= (c-a)(b-a)(c-b)(X^3 - (a+b+c)X+\cdots)
$$gives the result $$
(a+b+c)(c-a)(b-a)(c-b)
$$
