| (x+y)^2 zx zy |
| zx (z+y)^2 xy | = xyz(x+y+z)^3
| zy xy (z+x)^2 |
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Solution

$Let ∆ = |\begin{matrix} {(x + y)}^{2} & zx & yz \\ zx & {(y + z)}^{2} & xy \\ yz & xy & {(z + x)}^{2} \end{matrix}| Multiply R_{1} by z and R_{2} by x and R_{3} by y and dividing the determinant by xyz, we get ∆ = \frac{1}{xyz} |\begin{matrix} z {(x + y)}^{2} & z^{2} x & {yz}^{2} \\ {zx}^{2} & x {(y + z)}^{2} & x^{2} y \\ y^{2} z & {xy}^{2} & y {(z + x)}^{2} \end{matrix}| ∆ = \frac{xyz}{xyz} |\begin{matrix} {(x + y)}^{2} & z^{2} & z^{2} \\ x^{2} & {(y + z)}^{2} & x^{2} \\ y^{2} & y^{2} & {(z + x)}^{2} \end{matrix}| ∆ = |\begin{matrix} {(x + y)}^{2} & z^{2} & z^{2} \\ x^{2} & {(y + z)}^{2} & x^{2} \\ y^{2} & y^{2} & {(z + x)}^{2} \end{matrix}| Apply C_{1} = C_{1} - C_{3} and C_{2} = C_{2} - C_{3} ∆ = |\begin{matrix} (x + y + z) (x + y - z) & 0 & z^{2} \\ 0 & (x + y + z) (y + z - x) & x^{2} \\ (x + y + z) (y - z - x) & (x + y + z) (y - z - x) & {(z + x)}^{2} \end{matrix}| \Rightarrow ∆ = {(x + y + z)}^{2} |\begin{matrix} (x + y - z) & 0 & z^{2} \\ 0 & (y + z - x) & x^{2} \\ (y - z - x) & (y - z - x) & {(z + x)}^{2} \end{matrix}| Apply R_{3} = R_{3} - (R_{1} + R_{2}) ∆ = {(x + y + z)}^{2} |\begin{matrix} x + y - z & 0 & z^{2} \\ 0 & y + z - x & x^{2} \\ - 2 x & - 2 z & 2 zx \end{matrix}| = {(x + y + z)}^{2} [(x + y - z) \{2 zx (y + z - x) + 2 {zx}^{2}\} + z^{2} \{0 + 2 x (y + z - x)\}] = {(x + y + z)}^{2} [(x + y - z) \{2 zx (y + z - x + x)\} + 2 z^{2} x (y + z - x)] = {(x + y + z)}^{2} [(x + y - z) \{2 zx (y + z)\} + 2 z^{2} x (y + z - x)] = {(x + y + z)}^{2} \times 2 zx [(x + y - z) (y + z) + z (y + z - x)] = {(x + y + z)}^{2} \times 2 zx [xy + zx + y^{2} + yz - yz - z^{2} + yz + z^{2} - zx] = {(x + y + z)}^{2} \times 2 zx [xy + y^{2} + yz] = 2 xyz {(x + y + z)}^{3}$

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