[Math] Find the determinant without row expansion

Question

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.

Answer 1

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}