How to notice that $3^2 + (6t)^2 + (6t^2)^2$ is a binomial expansion. The other day during a seminar, in a calculation, a fellow student encountered this expression:
$$\sqrt{3^2 + (6t)^2 + (6t^2)^2}$$
He, without much thinking, immediately wrote down:
$$(6t^2+3)$$
What bothers me, is that I didn't see that. Although I suspected there might be some binomial expansion, I was too confused by the middle term to see it. Even after actually seeing the answer, it took me a good deal to realize what's happening.
My initial idea, by the way, was to substitute for $t^2$ and solve a quadratic equation. Horribly slow.
My questions is perhaps tackling a greater problem. I now see what happened, I can "fully" understand binomial expansions, yet I feel like I might end up encountering a similar expression and yet again not be able to see the result. My questions is thus:
How can I properly practice so I don't repeat this mistake?
This is perhaps too much of a soft-question, but I feel like many others have been in a position of understanding a problem, yet not feeling like they would be able to replicate it, and might actually have an answer that goes well beyond the obvious "just practice more".
Thanks for any input.

All you need to see is 
$$6^2=2 \cdot 3 \cdot 6$$
Then
$$\sqrt{3^2 + (6t)^2 + (6t^2)^2}=\sqrt{3^2 + 2 \cdot 3 \cdot (6t^2) + (6t^2)^2}$$
Anyhow, in general an expression like this is a monomial times a perfect square if and only if the product of the largest and smallest degree is exactly the square of half of the middle term. In this case, the monomial is exactly the smallest term in the expression.
In this case, a fast check if this is a perfect square would be to see that $3^2$ is a perfect square and :
$$3^2 \cdot (6t^2)^2 \stackrel{?}{=}  \left( \frac{(6t)^2}{2}\right)^2$$
