\(x^3+8=x^2-4\)
=>\(\left(x+2\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2-2x+4-x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2-3x+6\right)=0\)
mà \(x^2-3x+6=\left(x-\dfrac{3}{2}\right)^2+\dfrac{15}{4}>0\forall x\)
nên x+2=0
=>x=-2