PT$ \color{green}{\Leftrightarrow [(x+2)^2-1]+(x+3)^3+[(x+4)^4-1]=0}$$\color{green}{\Leftrightarrow (x+2+1)(x+2-1)+(x+3)^3+[(x+4)^2-1][(x+4)^2+1]=0}$
$\color{green}{\Leftrightarrow (x+3)(x+1)+(x+3)^3+(x+4+1)(x+4-1)(x^2+8x+17)=0}$
$\color{green}{\Leftrightarrow (x+3)[(x+1)+(x+3)^2]+(x+3)(x+5)(x^2+8x+17)=0}$
$\color{green}{\Leftrightarrow (x+3)(x^2+7x+10)+(x+3)(x+5)(x^2+8x+17)=0}$
$\color{green}{\Leftrightarrow (x+3)(x+5)(x+2)+(x+3)(x+5)(x^2+8x+17)=0}$
$\color{green}{\Leftrightarrow (x+3)(x+5)(x^2+9x+19)=0}$
Vậy phương trình có 4 nghiệm $\color{red}{x_1=-3,x_2=-5,x_{3,4}=\frac{-9\pm \sqrt{5}}{2}}$