Ta có: \(P+Q=0\)

Hay:\(P+\left(-x^2y^5+3y^3-3x^3+x^3y+2015\right)=0\)

\(\Rightarrow P=0-\left(-x^2y^5+3y^3-3x^3+x^3y+2015\right)\)

\(P=x^2y^5-3y^3+3x^3-x^3y-2015\)

\(D=\dfrac{16^3.3^{10}+120+6^9}{4^6.3^{12}+6^{11}}=\dfrac{4^3.4^3.\left(3^2\right)^5+120+6^9}{\left(4^3\right)^2.\left(3^2\right)^6+6^9.6^2}=\dfrac{4^3.4^3+120}{4^3.4^3.3^2}+\dfrac{6^9}{6^9.6^2}=\dfrac{120}{3^2}+\dfrac{1}{6^2}=\dfrac{40}{3}+\dfrac{1}{36}=\dfrac{481}{36}\approx13.36\left(1\right)\)