Evaluate$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&y&{x + y}\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right|$

$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&y&{x + y}\\y&{x + y}&x\\{x + y}&x&y\end{array}} \right|$

Applying ${C_1} \to {C_1} + {C_2} + {C_3}$,

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2(x + y)}&y&{x + y}\\{2(x + y)}&{x + y}&x\\{2(x + y)}&x&y\end{array}} \right|$

Taking $2(x + y)$common from ${C_1},$

we get
$2(x + y)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&y&{x + y}\\1&{x + y}&x\\1&x&y\end{array}} \right|$

Applying ${R_2} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_1},$

we get
$2(x + y)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&y&{x + y}\\0&x&{ - y}\\0&{x - y}&{ - x}\end{array}} \right|$

Expanding along ${C_1},$

we get
$2(x + y)[x( - x) - ( - y)(x - y)] = 2(x + y)[ - {x^2} + xy - {y^2})$

$= - 2(x + y)({x^2} - xy + {y^2}) = - 2({x^3} + {y^3})$