Solution

Applying \(R_1 \to xR_1, R_2 \to yR_2, R_3 \to zR_3\), we have \(\Delta = \frac{1}{xyz} |\begin{smallmatrix} x(y+z)^2 & x^2y & x^2z \\ xy^2 & y(x+z)^2 & y^2z \\ xz^2 & yz^2 & z(x+y)^2 \end{smallmatrix}|\) Taking common factors \(x, y, z\) from \(C_1, C_2, C_3\) respectively, we get \(\Delta = \frac{xyz}{xyz} |\begin{smallmatrix} (y+z)^2 & x^2 & x^2 \\ y^2 & (x+z)^2 & y^2 \\ z^2 & z^2 & (x+y)^2 \end{smallmatrix}|\) Applying \(C_2 \to C_2 - C_1, C_3 \to C_3 - C_1\), we have \(\Delta = |\begin{smallmatrix} (y+z)^2 & x^2-(y+z)^2 & x^2-(y+z)^2 \\ y^2 & (x+z)^2-y^2 & 0 \\ z^2 & 0 & (x+y)^2-z^2 \end{smallmatrix}|\) \(= |\begin{smallmatrix} (y+z)^2 & (x-y-z)(x+y+z) & (x-y-z)(x+y+z) \\ y^2 & (x+z-y)(x+z+y) & 0 \\ z^2 & 0 & (x+y-z)(x+y+z) \end{smallmatrix}|\) Taking \((x+y+z)^2\) common from \(C_2\) and \(C_3\), we get \(\Delta = (x+y+z)^2 |\begin{smallmatrix} (y+z)^2 & x-y-z & x-y-z \\ y^2 & x+z-y & 0 \\ z^2 & 0 & x+y-z \end{smallmatrix}|\) Applying \(R_1 \to R_1 - (R_2+R_3)\), we have \(\Delta = (x+y+z)^2 |\begin{smallmatrix} 2yz & -2z & -2y \\ y^2 & x+z-y & 0 \\ z^2 & 0 & x+y-z \end{smallmatrix}|\) \(= (x+y+z)^2 [2yz\{(x+z-y)(x+y-z)\} + 2z\{y^2(x+y-z)\} - 2y\{-z^2(x+z-y)\}]\) \(= (x+y+z)^2 [2yz(x^2 - (y-z)^2) + 2zy^2(x+y-z) + 2yz^2(x+z-y)]\) \(= (x+y+z)^2 [2yz(x^2 - y^2 + 2yz - z^2) + 2xy^2z + 2y^3z - 2y^2z^2 + 2xyz^2 + 2yz^3 - 2y^2z^2]\) \(= (x+y+z)^2 [2x^2yz - 2y^3z + 4y^2z^2 - 2yz^3 + 2xy^2z + 2y^3z - 2y^2z^2 + 2xyz^2 + 2yz^3 - 2y^2z^2]\) \(= (x+y+z)^2 [2x^2yz + 2xy^2z + 2xyz^2] = (x+y+z)^2 (2xyz)(x+y+z) = 2xyz(x+y+z)^3\)