\(x\left(x+3\right)^2-3x=\left(x+2\right)^3+1\\ \Rightarrow x\left(x^2+6x+9\right)-3x=x^3+9+6x\left(x+2\right)\\ =x^3+6x^2+6x=x^3+6x^2+12x+9\\ \Rightarrow12x+9=6x\\ \Rightarrow6x=-9\\ \Rightarrow x=-\dfrac{3}{2}\)
x( x + 3)2 - 3x = ( x + 2)3 + 1
=> x( x2 + 6x + 9) - 3x = x3 + 12x + 6x2 + 8 + 1
=> x3 + 6x2 + 9x - 3x = x3 + 12x + 6x2 + 9
=> x3 + 6x2 + 9x - 3x - x3 - 12x - 6x2 - 9 = 0
=> -6x - 9 = 0
=> -6x = 9
=> x = \(-\dfrac{9}{6}=-\dfrac{3}{2}\)
Vậy, phương trình nhận \(x=-\dfrac{3}{2}\) làm nghiệm duy nhất