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# Using properties of determinants, prove the following : $\begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3 \\ 1 & c^2+ab & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a^2+b^2+c^2).$

Toolbox:
• The value of the determinant can be found by expanding along any rows or columns.
• Elementary transformation in a determinant can be made by
• (a) Interchanging two rows or columns.
• (b) By adding or subtracting two or more rows or columns
Step 1:
Let $\Delta=\begin{vmatrix}1 & a^2+bc & a^3\\1 & b^2+ca&b^3\\1 & c^2+ab& c^3\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_3$ and $R_3 \rightarrow R_3-R_!$
$\Delta=\begin{vmatrix}1 & a^2+bc & a^3\\0 & b^2+ca-c^2-ab&b^3-c^3\\0 & c^2+ab-a^2-bc& c^3-a^3\end{vmatrix}$
$[a^3-b^3=(a-b)(a^2+ab+b^2)]$
This can be written as
$\Delta=\begin{vmatrix}1 & a^2+bc & a^3\\0 & (b+c)(b-c)+a(c-b)&(b-c)(b^2+bc+c^2)\\0 & (c+a)(c-a)+b(a-c)&(c-a)(c^2+ac+a^2)\end{vmatrix}$
Take (b-c) and (c-a) as factors from $R_2$ and $R_3$ respectively
$\Delta=(b-c)(c-a)\begin{vmatrix}1 & a^2+bc & a^3\\0 & b+c-a & b^2+bc+c^2\\0 & c+a-b & c^2+ac+a^2\end{vmatrix}$
Step 2:
$\Delta=(b-c)(c-a)\begin{vmatrix}1 & a^2+bc & a^3\\0 & -2(b-a) & (b^2-a^2)+c(b-a)\\0 & c+a-b & c^2+ac+a^2\end{vmatrix}$
Taking (a-b) as the common factor from $R_2$
$\Delta=(a-b)(b-c)(c-a)\begin{vmatrix}1 & a^2+bc & a^3\\0 & 2 & b+a+c\\0 & c+a-b & c^2+ac+a^2\end{vmatrix}$
Now expanding along $C_1$ we get
$\Delta=(a-b)(b-c)(c-a)[1[2(c^2+ac+a^2)-(c+a+b)(c+a-b)]]$
$\Delta=(a-b)(b-c)(c-a)[2(c^2+ac+a^2)-(c+a)^2-b^2)]$
On simplifying we get
$\Delta=(a-b)(b-c)(c-a)[2c^2+2ac+2a^2)-c^2-a^2-2ac+b^2]$
$\;\;\;=(a-b)(b-c)(c-a)[a^2+b^2+c^2]$
Hence $\Delta=(a-b)(b-c)(c-a)[a^2+b^2+c^2]$