# Prove that $\begin{vmatrix} 1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c).$

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^3 &b^3 & c^3\end{vmatrix}$

By applying $C_1\rightarrow C_1-C-2$ and $C_2\rightarrow C_2-C_3$

$\bigtriangleup=\begin{vmatrix}0 & 0 & 1\\(a-b) & (b-c) & c\\(a^3 -b^3)&(b^3-c^3) & c^3\end{vmatrix}$

But we know $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and $(b^3-c^3)=(b-c)(b^2+bc+c^2)$

Therefore $\bigtriangleup=\begin{vmatrix}0 & 0 & 1\\a-b & b-c & c\\(a-b)(a^2+ab+b^2) &(b-c)(b^2+bc+c^2) & c^3\end{vmatrix}$

Taking (a-b) as a common factor from $C_1$ and (b-c) from $C_2$ we get

$\bigtriangleup=(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\1 & 1 & c\\a^2+ab+b^2 &b^2+bc+c^2 & c^3\end{vmatrix}$

By applying $C_1\rightarrow C_1-C_2$

$\bigtriangleup=(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\1 & 1 & c\\a^2+ab-bc-c^2 &b^2+bc+c^2 & c^3\end{vmatrix}$

But $a^2+ab-bc-c^2$ can be written as $(a^2-c^2)+b(a-c)$

$\bigtriangleup=(a-b)(b-c)\begin{vmatrix}0 & 0 & 1\\0 & 1 & c\\(a^2-c^2)+b(a-c) &b^2+bc+c^2 & c^3\end{vmatrix}$

Now taking (a-c) as a common factor from $C_1$

$\bigtriangleup=(a-b)(b-c)(a-c)\begin{vmatrix}0 & 0 & 1\\0 & 1 & c\\a+c+b &b^2+bc+c^2 & c^3\end{vmatrix}$

Now expanding along $R_1$ we get,

Therefore $\bigtriangleup=(-1)(a-b)(b-c)(a-c)(a+b+c)$

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

Hence $\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^3 & b^3 & c^3\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)$

Hence proved