Question:
Suppose $t$ is a positive integer such that $\mathop{\text{lcm}}[12,t]^3=(12t)^2$. What is the smallest possible value for $t$?

Answer:
Recall the identity $\mathop{\text{lcm}}[a,b]\cdot \gcd(a,b)=ab$, which holds for all positive integers $a$ and $b$. Applying this identity to $12$ and $t$, we obtain $$\mathop{\text{lcm}}[12,t]\cdot \gcd(12,t) = 12t,$$and so (cubing both sides) $$\mathop{\text{lcm}}[12,t]^3 \cdot \gcd(12,t)^3 = (12t)^3.$$Substituting $(12t)^2$ for $\mathop{\text{lcm}}[12,t]^3$ and dividing both sides by $(12t)^2$, we have $$\gcd(12,t)^3 = 12t,$$so in particular, $12t$ is the cube of an integer. Since $12=2^2\cdot 3^1$, the smallest cube of the form $12t$ is $2^3\cdot 3^3$, which is obtained when $t=2^1\cdot 3^2 = 18$. This tells us that $t\ge 18$.

We must check whether $t$ can be $18$. That is, we must check whether $\mathop{\text{lcm}}[12,18]^3=(12\cdot 18)^2$. In fact, this equality does hold (both sides are equal to $6^6$), so the smallest possible value of $t$ is confirmed to be $\boxed{18}$.