Gọi (x-1) =a; (x+3) = b
=> (2x+2) = (a+b)
Ta có: \(a^3+b^3=\left(a+b\right)^3\)
=> \(a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)
=> \(3ab\left(a+b\right)=0\)
- TH1: a=0 => x-1=0 => x=1
- TH2: b=0 => x+3 =0 => x= -3
- TH3: a+b=0 => 2x+2=0 => x=-1
KL: \(x\in\left\{1;-3;-1\right\}\)