Show that $a+b+c$ is a factor of $$\det \begin{pmatrix} b+c & a & a^3\\ c+a & b & b^3\\ a+b & c & c^3 \end{pmatrix}$$ and express the determinant as a product of five factors.
I was able to get the first phase of the question but the second part proved difficult for me. The answer is $$(a+b+c)^2(a-b)(b-c)(c-a)$$ but how to get it remains a mystery to me.
Similar problem like this was given me. Here I just want to be sure if am correct. Express $\det \begin{bmatrix}1 & 1 & 1\\ a^2 & b^2 & c^2\\ [a+b]^2 & [c+a]^2 & [a+b]^2\end{bmatrix}$ as a product of linear factors After solving on my own and following the guide provided here. I got $(b-a)(b+a)(b-c)(2a+b+c)$. However, I was meant to understand that the correct solution to the question is $2(a-b)(b-c)(a+b+c)$ So which is the correct answer? Mine or the latter?
$$\begin{array}{ccc} \det\left(\begin{array}{ccc}b+c&a&a^3\\c+a&b&b^3\\ a+b&c&c^3\\ \end{array} \right) &\stackrel{c_2+c_1\to c_1}{=} & \det\left(\begin{array}{ccc}a+b+c&a&a^3\\a+b+c&b&b^3\\ a+b+c&c&c^3& \end{array} \right)\\ \\ &=& (a+b+c)\cdot \det\left(\begin{array}{ccc}1&a&a^3\\1&b&b^3\\ 1&c&c^3& \end{array} \right)\\ \\ &\stackrel{-r_2+r_1 \to r_1}{=}& (a+b+c)\cdot \det\left(\begin{array}{ccc}0&a-b&a^3-b^3\\1&b&b^3\\ 1&c&c^3& \end{array} \right)\\ \\ &=&(a+b+c)\cdot(a-b)\cdot \det\left(\begin{array}{ccc}0&1&a^2+ab+b^2\\1&b&b^3\\ 1&c&c^3& \end{array} \right)\\ \\ &\stackrel{-r_3+r_2\to r_2}{=}& (a+b+c)\cdot (a-b)\cdot \det\left(\begin{array}{ccc}0&1&a^2+ab+b^2\\0&b-c&b^3-c^3\\ 1&c&c^3& \end{array} \right)\\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot \det\left(\begin{array}{ccc}0&1&a^2+ab+b^2\\0&1&b^2+bc+c^2\\ 1&c&c^3& \end{array} \right)\\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot((b^2+bc+c^2)-(a^2+ab+b^2)) \\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot (bc+c^2-a^2-ab) \\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot ((c-a)(c+a)+b(c-a))\\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot ((c-a)(c+a+b)) \\ \\ &=& (a+b+c)^2\cdot(a-b)\cdot(b-c)\cdot(c-a) \end{array}$$