x + y + z = 0
=> x + y = -z
<=> (x + y)^3 = (-z)^3
<=> x^3 + 3x^2y + 3xy^2 + y^3 = -z^3
<=> x^3 + y^3 + z^3 = -3x^2y - 3xy^2
<=> x^3 + y^3 + z^3 = -3xy(x+y)
<=> x^3 + y^3 + z^3 = -3xy(-z)
<=> x^3 + y^3 + z^3 = 3xyz
<=>\(\dfrac{x^3+y^3+z^3}{xyz}=3\Leftrightarrow\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=3\Leftrightarrow P=3\)