Ta có: \(a+\dfrac{1}{b}=-4\)
\(\Rightarrow\left(a+\dfrac{1}{b}\right)^3=\left(-4\right)^3\)
\(\Rightarrow a^3+3.a^2.\dfrac{1}{b}+3.a.\dfrac{1}{b^2}+\dfrac{1}{b^3}=-64\)
\(\Rightarrow a^3+\dfrac{3a^2}{b}+\dfrac{3a}{b^2}+\dfrac{1}{b^3}=-64\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-\dfrac{3a^2}{b}-\dfrac{3a}{b^2}\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-\dfrac{3a}{b}\left(a+\dfrac{1}{b}\right)\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-3.\left(-4\right).\left(-4\right)\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-112\)
\(P=a^3+\dfrac{1}{b^3}\\ =\left(a+\dfrac{1}{b}\right)\left(a^2+\dfrac{a}{b}+\dfrac{1}{b^2}\right)\\ =-4\left(a^2+\dfrac{2a}{b}+\dfrac{1}{b^2}-\dfrac{a}{b}\right)\\ =-4\left[\left(a+\dfrac{1}{b}\right)^2-\dfrac{a}{b}\right]\\ =-4\left[\left(-4\right)^2-\left(-4\right)\right]\\ =-80\)
\(P=a^3+\dfrac{1}{b^3}\\ =\left(a+\dfrac{1}{b}\right)\left(a^2-ab+\dfrac{1}{b^2}\right)\\ =\left(a+\dfrac{1}{b}\right)\left(a^2+2ab+\dfrac{1}{b^2}-3ab\right)\\ =\left(a+\dfrac{1}{b}\right)\left[\left(a+\dfrac{1}{b}\right)^2-3ab\right]\\ =-4\left[\left(-4\right)^2-3\left(-4\right)\right]=-112\)
