$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$

$\Leftrightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}$

$\Leftrightarrow \frac{x+y}{xy}=\frac{z-(x+y+z)}{z\left( x+y+z \right)}$

$\Leftrightarrow \frac{x+y}{xy}=\frac{-\left( x+y \right)}{z(x+y+z)}$

$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$

$\Leftrightarrow \left( x+y \right)\left( \frac{1}{xy}+\frac{1}{z\left( x+y+z \right)} \right)=0$

$\Leftrightarrow \left( x+y \right)\left( \frac{xz+zy+{{z}^{2}}+xy}{xyz\left( x+y+z \right)} \right)=0$

$\Leftrightarrow \left( x+y \right).\frac{(x+z)(z+y)}{xyz\left( x+y+z \right)}=0$

$\Leftrightarrow \frac{(x+y)(x+z)(z+y)}{xyz\left( x+y+z \right)}=0$

Suy ra $x+y=0$ hoặc $x+z=0$ hoặc $z+y=0$

$P=\left( {{x}^{25}}+{{y}^{25}} \right).\left( {{y}^{3}}+{{z}^{3}} \right).\left( {{z}^{2006}}-{{x}^{2006}} \right)$

Trường hợp 1 : $x+y=0$ $x=-y\Rightarrow {{x}^{25}}=\left( -{{y}^{25}} \right)=-{{y}^{25}}\Rightarrow {{x}^{25}}+{{y}^{25}}=0\Rightarrow P=0$

Tương tự trường hợp 2 và 3 tính được P = 0