\(x=\dfrac{4a+4b}{3\left(a^3-b^3\right)}\) ; \(y=\dfrac{3\left(a-b\right)^2}{4\left(a+b\right)}\)
\(\Rightarrow P=xy=\dfrac{4\left(a+b\right)}{3\left(a^3-b^3\right)}.\dfrac{3\left(a-b\right)^2}{4\left(a+b\right)}=\dfrac{\left(a-b\right)^2}{\left(a-b\right)\left(a^2+ab+b^2\right)}=\dfrac{a-b}{a^2+ab+b^2}\)