Δ = $\begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix}$ Applying $C_1$ → $C_1+C_2+C_3$ Δ = $\begin{vmatrix} a+b+c & b & c \\ 0 & b-c & c-a \\ 2(a+b+c) & c+a & a+b \end{vmatrix}$ Δ = (a + b + c) $\begin{vmatrix} 1 & b & c \\ 0 & b-c & c-a \\ 0 & c+a-2b & a+b-2c \end{vmatrix}$ $R_3$ → $R_3-2R_1$ Δ = (a + b + c) $\begin{vmatrix} 1 & b & c \\ 0 & b-c & c-a \\ 0 & c+a-2b & a+b-2c \end{vmatrix}$ Expanding along $C_1$, we have, Δ = (a +b +c) ((b –c) (a + b – 2c) – (c – a) (c + a – 2b)) ⇒ Δ = (a + b + c) ((ba + $b^2$ - 2bc - ca - cb + $2c^2$ - ($c^2$ + ac - 2bc - ac - $a^2$ + 2ab)) ⇒ Δ = (a + b + c) ($a^2+b^2+c^2$ - ca - bc - ab) ⇒ Δ = (a + b + c) ($a^2+b^2+c^2$ - ab - bc - ac) ⇒ Δ = $a^3+b^3+c^3$ - 3abc = R.H.S.