Bài 1:
a: \(x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\cdot x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)\)
\(=3^3-3\cdot3=27-9=18\)
b: x^2+1/x^2=14
=>(x+1/x)^2-2=14
=>(x+1/x)^2=16
=>x+1/x=4 hoặc x+1/x=-4
TH1: x+1/x=4
\(A=\left(x+\dfrac{1}{x}\right)^3-3\cdot x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\)
\(=4^3-3\cdot4=64-12=52\)
TH2: x+1/x=-4
\(A=\left(-4\right)^3-3\cdot\left(-4\right)=-64+12=-52\)