a, P xác định khi \(x^3-8\ne0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\ne0\)
\(\Leftrightarrow x\ne2\left(\text{Vì }x^2+2x+4>0\right)\)
b, \(P=\dfrac{3x^2+6x+12}{x^3-8}=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
c, \(x=\dfrac{4001}{2000}\Rightarrow P=\dfrac{3}{\dfrac{4001}{2000}-2}=6000\)