# 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.

• Even though your determinant is perfectly clear, there will probably be some complaints because you didn't format it using Latex commands. Personally, I'm much more concerned about the lack of periods and upper-case letters. :-) – bubba Mar 21 '14 at 1:06
• It is the first time that I use this website so I didn't know how to do it right – Ahmad Amr Ebeid Mar 21 '14 at 1:10
• You don't know how to type a period?? :-) I'm not complaining about the math formatting, but I expect someone will. Or, some nice person might even fix it for you. – bubba Mar 21 '14 at 1:12
• Anyway, back to the mathematics. After you factored out $(a-b)(b-c)(c-a)$, what determinant did you have left? – bubba Mar 21 '14 at 1:14

\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 $$(a+b+c)(c-a)(b-a)(c-b)$$