Every now and then we all come across an algebra problem which we have no idea how to go about. We've all been there: it seems really, really simple, but you just can't wrap your head around solving it or finding the desired expression.
Usually some really epic trick is used which takes the problem from $$\text{ unsolved, still wandering } \longrightarrow \text{ almost solved, finishing details}$$
An example of this is the following problem, which is 2019 AIME I P8:
It may seem quite confusing how to approach this at first. We can't really exponentiate the powers of $10$ to take them directly to the powers of $12$, but we do notice that the exponents differ by $2$, which gives us inspiration to multiply the given expression by $\sin^2 x + \cos^2 x$, which is again really just multiplying by $1$ so we don't need to translate anything back. Multiplying, we see that $$(\sin^{10}x + \cos^{10}x)(\sin^2x + \cos^2x) = \sin^{12}x + \cos^{12}x + 2\sin^{10}x \cos^2 x $$ It seems like we're going in the right direction, but there's that pesky $2\sin^{10}x \cos^2 x$ term which we don't know how to deal with. But this is just equal to $(\sin^2x \cos^2x)(\sin^{8}x + \cos^{8}x)$, which motivates us to find $\sin^{8}x + \cos^{8}x$, which takes us back to the original problem, just with a lesser exponent?! This motivates a recursion-styled approach, and at this point I guess you could say that you've crossed the main bridge connecting this problem to its solution. Repeating this, your final solution looks something like the following:
Solution:
We perform a recursion style solution on this problem, relying on the fact that $\sin^2x + \cos^2x = 1$Note that $$(\sin^2x+\cos^2x)(\cancel{\sin^2x+\cos^2x}) = \sin^4x + \cos^4x + 2 \sin^2x \cos^2x = 1 \implies \sin^4x + \cos^4x = 1 - 2 \sin^2x \cos^2x$$Then, $$(\sin^4x + \cos^4x)(\cancel{\sin^2x + \cos^2x}) = \sin^6x + \cos^6x + \sin^4x \cos^2x + \sin^2x \cos^4x = \sin^6x + \cos^6x + (\sin^2x \cos^2x)(\cancel{\sin^2x + \cos^2x}) = 1 - 2 \sin^2x \cos^2x \implies \sin^6x + \cos^6x = 1 - 3 \sin^2x \cos^2x$$Then, $$(\sin^6x + \cos^6x)(\cancel{\sin^2x + \cos^2x}) = \sin^8x + \cos^8x + \sin^6x \cos^2x + \sin^2x \cos^6x = \sin^8x + \cos^8x + (\sin^2x \cos^2x)(\sin^4x + \cos^4x) = \sin^8x + \cos^8x + (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x) = 1 - 3 \sin^2x \cos^2x \implies \sin^8x + \cos^8x = 1 - 3 \sin^2x \cos^2x - (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x)$$Then, $$(\sin^8x + \cos^8x)(\cancel{\sin^2x + \cos^2x}) = \sin^{10}x + \cos^{10}x + \sin^8x \cos^2x + \sin^2x \cos^8x = \sin^{10}x + \cos^{10}x + (\sin^2x \cos^2x)(\sin^6x + \cos^6x) = \sin^{10}x + \cos^{10}x + (\sin^2x \cos^2x)(1 - 3 \sin^2x \cos^2x) = 1 - 3 \sin^2x \cos^2x - (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x)$$At this point, we know that the value of $\sin^{10}x + \cos^{10}x = \frac{11}{36}$, so letting $y = \sin^2x \cos^2x$ and plugging these values into the above equation, we get $y=\frac{1}{6}$ and $y=\frac{5}{6}$, but we can weed out the second one because it gives a negative value for the computation of $\sin^4x + \cos^4x$, which is clearly not possible.Now, we apply the recursion one last time, to find the desired term.We have $$(\sin^{10}x+\cos^{10}x)(\cancel{\sin^2x+\cos^2x}) = \sin^{12}x+\cos^{12}x + (\sin^2x \cos^2x)(\sin^8x + \cos^8x) = \sin^{12}x+\cos^{12}x + (\sin^2x \cos^2x)(1 - 3 \sin^2x \cos^2x - (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x)) = \sin^{12}x + \cos^{12}x + (a)(1-3a-(a)(1-2a)) = \frac{11}{36}$$Solving this equation for $\sin^{12}x + \cos^{12}x$, we obtain $\boxed{\frac{13}{54}}$.
I think at least at the AMC/AIME level, algebra problems fall into one of the following categories:
1) given an expression with two or more variables (with some possibly integer constraint), find an expression using the same variables.
2) polynomials (this mainly consists of manipulations with roots of the polynomials with vieta's or taking advantage of the "integer"-type constraints)
3) complex numbers (this one is so unique that I think it deserves its own category)
In the first 10 of AIME, I used to struggle the most with the first category. A key intuition that I picked up while solving tons of problem, is, that there are in general two ends of the problem as illustrated in the following picture, specifically in the context of the problem we just went over:
However, it isn't as simple as that because such a manipulation can be hard to think of, especially if you limit yourself to only what you have and what you need to find. This is what really went through my mind while solving this problem (for brevity, I have written $f(n) = \sin^nx + \cos^nx$), which I guess you could call thinking "outside of the box" (which, I think is quite important in AIME algebra problems): 
In these type of problems, it's important to think about where the main condition lies. Does it lie in the specific value of the numbers $10$ and $12$? Or is only the difference of the two powers relevant, which is $2$? Naturally checking through the ideas gives an intuition that the latter is correct.
AIME Algebra problems can actually get quite perplexing at times (especially the later ones), but I think the "thinking outside the problem space to move into the solution space" approach and finding where exactly the main difficulty of the problem lies (in this case, the difference of powers) is quite crucial.
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