so for AM=GM
$\frac{x1+x2+x3}{3}=\sqrt[3]{x1*x2*x3}$
now we can define range $0\leq(x1*x2*x3)\leq64$
so GM can fall in range of $0\leq\sqrt[3]{(x1*x2*x3)}\leq4$
also x1*x2*x3 $\in$ Integer
also $0\leq\frac{x1+x2+x3}{3}\leq4$
possible value of $\sqrt[3]{x1*x2*x3}$ can be (0,1,2,3,4) = range of $\frac{x1+x2+x3}{3}$
only triplets that satisfy both condition are (x1,x2,x3)={(0,0,0),(1,1,1),(2,2,2),(3,3,3),(4,4,4)}
now probability of AM=GM = $\frac{5}{5^{3}}$=$\frac{1}{5^{2}}$
Thus option A
One can verify
#include<stdio.h>
int main(){
int i,j,k,sum,mul;
for(i=0;i<5;i++)
for(j=0;j<5;j++)
for(k=0;k<5;k++){
sum=i+j+k;
sum=sum*sum*sum;
mul=27*i*j*k;
if (sum==mul)
printf("%d %d %d\n",i,j,k);
}
return 0;
}
