Is it difficult?

Rearranging terms, we get: $x^3+(x+1)^3=(x+3)^3-(x+2)^3$ $\Rightarrow (2x+1)(x^2+x+1)=(3x^2+15x+19)$ $\Rightarrow x^3-6x-9=0$ $\Rightarrow (x-3)(x^2+3x+3)=0$ $\text{ Hence only real x=}\color{magenta}{3}$

put y=x+3,substitute in expression (x+3-3)^3+(x+3-2)^3+(x+3-1)^3=x^3 ,so. ,,,,,(y-3)^3+(y-2)^3+(y-1)^3=y^3 .by solving expression result 2y^3-9y^2+21y-18=0.(y^2-3y+3)(y-6)=0,,y=6 or y=3+or-√9-4×3(not applicable ),y=6,,,,,X=6-3=3####