设三个数为x,y,z,则\text{设三个数为}x,y,z,则

x+y+z≥3(x⋅y⋅z)13＝3a13x+y+z\ge 3\left( x\cdot y\cdot z \right) ^{\frac{1}{3}}＝3a^{\frac{1}{3}}
故最小值为a13\text{故最小值为}a^{\frac{1}{3}}

次方直接用pow次方直接用pow

include <bits/stdc++.h>
using namespace std;
int main()
{
    double n;
    scanf("%lf", &n);
    double m = pow(n, 1.0 / 3);
    printf("%.3lf\n", m + m + m);
    return 0;
}