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.

The radicand is obviously a multiple of $9$.  It factors to $9(1+(2t)^2+(2t^2)^2)$, and that simplifies to $9(1+4t^2+4t^4)$.  It is usual to write polynomials in decreasing order of powers, so you get $9(4t^4+4t^2+1)$, which equals $9(4(t^2)^2+4t^2+1)$.  Someone with experience in factoring things may recognize the last expression as $9(2t^2+1)^2$ (you have something of the form "$4x^2+4x+1$").  When you take the square root of that, you get $\sqrt{9}\sqrt{(2t^2+1)^2}=3(2t^2+1)=6t^2+3$.
This is not the fastest way to do it.  It just takes practice.  I think it is helpful here to factor out the $9$ to simplify and recognize the radicand as a quadratic in $t^2$, so you are really factoring a quadratic.
