#include <bits/stdc++.h>
using namespace std;

#define vt vector
#define um unordered_map
#define pii pair
#define mp map

#define ff first
#define ss second
#define pb push_back
#define sz(x) x.size()

#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()

#define no cout << "NO" << nl
#define yes cout << "YES" << nl
#define nl "\n"

typedef long long ll;
typedef double db;

const int MAX = 1e9+65;
const int MIN = -(1e9+65);
const int MOD = 1e9+7;
const int N = 2*1e5;
const int NLL = 1e9;
const int N6 = 1e6;

void solve(){
    int a,b,c; cin >> a >> b >> c;
    int p1 = b/__gcd(a,b);
    int p2 = c/__gcd(b,c);
    int p3 = a/__gcd(c,a);
    ll l1 = p1*p2/__gcd(p1*p2);
    ll l2 = p3*l1/__gcd(1LL*p3,l1);
    cout << l2 << nl;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t; cin >> t;while(t--) solve();
    //solve();
    return 0;
}

$$$ x = lcm(\frac{b}{gcd(a, b)}, \frac{c}{gcd(b, c)},\frac{a}{gcd(a, c)}) $$$

First consider the following statements: \begin{align} ax \equiv 0 \pmod b \end{align} \begin{align} bx \equiv 0 \pmod c \end{align} \begin{align} cx \equiv 0 \pmod a \end{align}

This means that $$$b$$$ is a multiple of $$$ax$$$, $$$c$$$ is a multiple of $$$bx$$$ and $$$a$$$ is a multiple of $$$cx$$$. We can write it as:

\begin{align} ax \mid b \end{align} \begin{align} bx \mid c \end{align} \begin{align} cx \mid a \end{align}

($$$ a \mid b $$$ means $$$a$$$ divides $$$b$$$)

There is a known property in number theory: \begin{align} m \mid nx \qquad \Longleftrightarrow \qquad \frac{m}{\gcd(m,n)} \mid x \end{align}

Hence above expressions can be simplified as: \begin{align} \frac{b}{gcd(a, b)} \mid x \end{align} \begin{align} \frac{c}{gcd(b, c)} \mid x \end{align} \begin{align} \frac{a}{gcd(c, a)} \mid x \end{align}

The expression basically tells what $$$x$$$ must contribute (in prime factors term) to make b divisible by $$$ax$$$. Let: \begin{align} d_1 = \frac{b}{gcd(a, b)} \end{align} \begin{align} d_2 = \frac{c}{gcd(b, c)} \end{align} \begin{align} d_3 = \frac{a}{gcd(c, a)} \end{align} Then we get: \begin{align} d_1 \mid x , \qquad d_2 \mid x , \qquad d_3 \mid x \end{align}

Which basically means that $$$x$$$ is multiple of all $$$d_1, d_2, and\; d_3$$$. Since we are looking for smallest value, we are looking for smallest such $$$x$$$ that is still a multiple of $$$d_1, d_2, and\; d_3$$$. Now by definition: \begin{align} LCM(a,b,c) = Smallest/Least\;common\;multiple\;of\;a,\;b\;and\;c \end{align}

Hence we arrive to the answer of: \begin{align} x = lcm(\frac{b}{gcd(a, b)}, \frac{c}{gcd(b, c)},\frac{a}{gcd(a, c)}) \end{align}