# Prove that $\begin{vmatrix}bc - a^2 & ca - b^2 & ab - c^2\\ca - b^2 & ab - c^2 & bc - a^2\\ab -c^2 & bc - a^2 & ca - b^2\end{vmatrix}$ is divisible by a+b+c and find the quotient.

Toolbox:
• If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
• $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
• If some or all elements of a row or column of a determinant are expressed as a sum of two or more terms,then the determinant can be expressed as sum of two (or more) determinants.
• If each row or column of a determinant is multiplied by k,then its value is multiplied by k.
Let $\Delta=\begin{vmatrix}bc - a^2 & ca - b^2 & ab - c^2\\ca - b^2 & ab - c^2 & bc - a^2\\ab -c^2 & bc - a^2 & ca - b^2\end{vmatrix}$
Apply $R_1\rightarrow R_1+R_2+R_3$
$\Delta=\begin{vmatrix}ab+bc+ca-a^2-b^2-c^2 & ab+bc+ca-a^2-b^2-c^2 & ab+bc+ca-a^2-b^2-c^2 \\ca - b^2 & ab - c^2 & bc - a^2\\ab -c^2 & bc - a^2 & ca - b^2\end{vmatrix}$
Take $(ab+bc+ca-a^2-b^2-c^2)$ as the common factor from $R_1$
$\Delta=ab+bc+ca-a^2-b^2-c^2\begin{vmatrix}1 & 1 & 1\\ca - b^2 & ab - c^2 & bc - a^2\\ab -c^2 & bc - a^2 & ca - b^2\end{vmatrix}$
Apply $C_2\rightarrow C_2-C_3$ and $C_3\rightarrow C_3-C_1$
$\Delta=ab+bc+ca-a^2-b^2-c^2\begin{vmatrix}1 & 0 & 0\\ca - b^2 & ab-bc+a^2 - c^2 & bc-ca+b^2 - a^2\\ab -c^2 & bc-ca+b^2 - a^2 & ca-ab+c^2 - b^2\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=(ab+bc+ca-a^2-b^2-c^2)[ab-bc+a^2-c^2)(ca-ab+c^2-b^3)-(bc-ca+b^2-a^2)(bc-ca+b^2-a^2)]$
On simplifying we get,
$(ab+bc+ca-a^2-b^2-c^2)[b(a-c)+(a+c)(a-c)]\times [a(c-b)+(c-b)(c+b)]-[c(b-a)+(b-a)(b+a)]\times [c(b-a)+(b-a)(b+a)]$
Taking the common factors we get,
$(ab+bc+ca-a^2-b^2-c^2)[(a-c)(a+b+c)((c-b)(a+b+c)]-[(b-a)(a+b+c)(b-a)(a+b+c)]$
Again taking $(a+b+c)^2$ as the common factor
$(a+b+c)^2(ab+bc+ca-a^2-b^2-c^2)[(c-a)(c-b)-(b-a)(b-a)]$
Since it is given (a+b+c) is a factor
The given determinant is divisible by (a+b+c)
The quotient is
$[ab+bc+ca-b^2-b^2-c^2][ab+bc+ca-a^2-b^2-c^2]$
$\;\;=(ab+bc+ca-a^2-b^2-c^2)^2$