Consider the function $f(x)=\frac{x^3}{e^x}$ $f'(x)={ e }^{ -x }(3-x){ x }^{ 2 }$ So the function is increasing in the interval $(-\infty,3)$ and decreasing in the interval $(3,\infty)$ $f(x)=0$ at $x=0$ Also calculate $f''(x)$ and we find the global maximum of the function $f(x)$ is at $x=3$ and $f(3)=\frac{27}{e^3}$ Now we can conclude that (1) If $a>\frac{27}{e^3}$ then there is no solution (2) If $a=\frac{27}{e^3}$ then there is 1 solution (3) If $0<a<\frac{27}{e^3}$ then there are 2 solutions (4) If $a<0$ then there is 1 solution