# Using properties of determinates prove that

To prove:

$\begin{vmatrix} a-b &b+c &a \\ b-c&c+a &b \\ c-a& a+b &c \end{vmatrix} =a^{3}+b^{3}+c^{3}-3abc$

$LHS =\begin{vmatrix} a-b &b+c &a \\ b-c&c+a &b \\ c-a& a+b &c \end{vmatrix}$

$C_{2} \rightarrow C_{2}+C_{3}$

$LHS =\begin{vmatrix} a-b &a+b+c &a \\ b-c&c+a+b &b \\ c-a& a+b+c &c \end{vmatrix}$

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

$\\R_{1} \rightarrow R_{1}-R_{2} \\ R_{2} \rightarrow R_{2}-R_{3} \\$

$LHS =(a+b+c)\begin{vmatrix} a-2b+c &0 &a-b \\ b-2c+a&0&b-c \\ c-a& 1 &c \end{vmatrix}$

$LHS =(a+b+c)[-1( (a-2b+c)(b-c) - (a-b)( b-2c+a))]$

$LHS =(a+b+c)[-1( ab-2b^2+bc -ac+2bc-c^2 - ab +2ac -a^2+ b^2-2bc+ab)]$

$LHS =(a+b+c)[a^2+b^2+c^2-ab-bc-ac]$

$LHS =a^3+b^3+c^3-3abc$

$LHS =RHS$

Hence proved

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