Binomial coefficient after expansion

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I am trying to solve an exercise where in the final step, I need to find the coefficient of $x^7y^5$ in $(x+y)^{12} + 7(x^2+y^2)^6 + 2(x^3+y^3)^4 + 2(x^4+y^4)^3 + 2(x^6+y^6)^2 + 4(x^{12}+y^{12}) + 6(x+y)^2(x^2+y^2)^5$.

To proceed I first removed the terms like $4(x^{12}+y^{12})$ since they would not be involved in the final answer but don't know how to proceed further to reach the final answer. Any Help would be appreiated.

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I think the answer is hidden inside 2 terms only, $(x+y)^{12}$ and $6(x+y)^2$$(x^2 + y^2)^5$, because if you think about it, terms like $7(x^2+y^2)^6$ can't possibly have a $x^7y^5$, since power of $x$ and $y$ increases by 2. After that it's pretty easy to partially expand( as in just find the terms you need via binomial expansion, keeping in mind there are several ways to get $x^7y^5$ from both terms).

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The answer is hidden inside 2 terms only, (x+y)12 and 6(x+y)2(x2+y2)5, because if you think about it, terms like 7(x2+y2)6 can't possibly have a x7y5, since power of x and y increases by 2. After that it's pretty easy to partially expand( as in just find the terms you need via binomial expansion, keeping in mind there are several ways to get x7y5 from both terms).