Divide cuboid into cubes such that sum of volumes is maximum

Given the length, breadth, height of a cuboid. The task is to divide the given cuboid in minimum number of cubes such that size of all cubes is same and sum of volumes of cubes is maximum. 
Examples: 

Input : l = 2, b = 4, h = 6
Output : 2 6
A cuboid of length 2, breadth 4 and 
height 6 can be divided into 6 cube 
of side equal to 2.
Volume of cubes = 6*(2*2*2) = 6*8 = 48.
Volume of cuboid = 2*4*6 = 48.

Input : 1 2 3
Output : 1 6

First of all, we are not allowed to waste volume of cuboid as we need maximum volume sum. So, each side should be completely divide among all cubes. And since each of three side of cubes are equal, so each side of the cuboid need to be divisible by same number, say x, which will going to be the side of the cube. So, we have to maximize this x, which will divide given length, breadth and height. This x will be maximum only if it is greatest common divisor of given length, breadth and height. So, the length of the cube will be GCD of length, breadth and height.
Now, to compute number of cubes, we know total volume of cuboid and can find volume of one cube (since side is already calculated). So, total number of cubes is equal to (volume of cuboid)/(volume of cube) i.e (l * b * h)/(x * x * x). 
Below is implementation of this approach: 

C++

// CPP program to find optimal way to break
// cuboid into cubes.
#include <bits/stdc++.h>
using namespace std;

// Print the maximum side and no of cube.
void maximizecube(int l, int b, int h)
{
    // GCD to find side.
    int side = __gcd(l, __gcd(b, h));

    // dividing to find number of cubes.
    int num = l / side;
    num = (num * b / side);
    num = (num * h / side);

    cout << side << " " << num << endl;
}

// Driver code
int main()
{
    int l = 2, b = 4, h = 6;

    maximizecube(l, b, h);
    return 0;
}

Java

// JAVA Code for Divide cuboid into cubes
// such that sum of volumes is maximum
import java.util.*;

class GFG {

    static int gcd(int m, int n)
    {
        if(n == 0)
            return m;
        else if(n > m)
            return gcd(n,m);
        else
            return gcd(n, m % n);
    }

    // Print the maximum side and no
    //     of cube.
    static void maximizecube(int l, int b,
                                    int h)
    {
        // GCD to find side.
        int side = gcd(l, gcd(b, h));

        // dividing to find number of cubes.
        int num = l / side;
        num = (num * b / side);
        num = (num * h / side);

       System.out.println( side + " " + num);
    }

    /* Driver program  */
    public static void main(String[] args)
    {
         int l = 2, b = 4, h = 6;
         maximizecube(l, b, h);
    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 code to find optimal way to break
# cuboid into cubes.
from fractions import gcd

# Print the maximum side and no of cube.
def maximizecube( l , b , h ):

    # GCD to find side.
    side = gcd(l, gcd(b, h))

    # dividing to find number of cubes.
    num = int(l / side)
    num = int(num * b / side)
    num = int(num * h / side)

    print(side, num)

# Driver code
l = 2
b = 4
h = 6

maximizecube(l, b, h)

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# Code for Divide cuboid into cubes
// such that sum of volumes is maximum
using System;

class GFG {

    static int gcd(int m, int n)
    {
        if(n == 0)
            return m;
        else if(n > m)
            return gcd(n,m);
        else
            return gcd(n, m % n);
    }

    // Print the maximum side and no
    // of cube.
    static void maximizecube(int l, int b,
                                    int h)
    {
        // GCD to find side.
        int side = gcd(l, gcd(b, h));

        // dividing to find number of cubes.
        int num = l / side;
        num = (num * b / side);
        num = (num * h / side);

    Console.WriteLine( side + " " + num);
    }

    /* Driver program */
    public static void Main()
    {
        int l = 2, b = 4, h = 6;
        maximizecube(l, b, h);
    }
}

// This code is contributed by vt_m.

Javascript

<script>

// JavaScript program for Divide cuboid into cubes
// such that sum of volumes is maximum
    function gcd(m, n)
    {
        if(n == 0)
            return m;
        else if(n > m)
            return gcd(n,m);
        else
            return gcd(n, m % n);
    }

    // Print the maximum side and no
    //     of cube.
    function maximizecube(l, b, h)
    {

        // GCD to find side.
        let side = gcd(l, gcd(b, h));

        // dividing to find number of cubes.
        let num = l / side;
        num = (num * b / side);
        num = (num * h / side);

       document.write( side + " " + num);
    }

// Driver code
        let l = 2, b = 4, h = 6;
         maximizecube(l, b, h);

         // This code is contributed by sanjoy_62.
</script>

PHP

<?php
// PHP program to find optimal way
// to break cuboid into cubes.

// Recursive function to
// return gcd of a and b
function __gcd($a, $b)
{

    // Everything divides 0
    if($a == 0 or $b == 0)
        return 0 ;

    // base case
    if($a == $b)
        return $a ;

    // a is greater
    if($a > $b)
        return __gcd($a - $b , $b ) ;

    return __gcd($a , $b - $a) ;
}


// Print the maximum side and no of cube.
function maximizecube($l, $b, $h)
{

    // GCD to find side.
    $side = __gcd($l, __gcd($b, $h));

    // dividing to find number of cubes.
    $num = $l / $side;
    $num = ($num * $b / $side);
    $num = ($num * $h / $side);

    echo $side , " " , $num ;
}

    // Driver code
    $l = 2;
    $b = 4;
    $h = 6;
    maximizecube($l, $b, $h);

// This code is contributed by anuj_67.
?>

Output: 

2 6

Time Complexity: O(log₂n), where n is the upper limit of b and h.
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.