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Update thanks to comments fun = A^2 + A B + B^2 == (a^2 + a b + b^2) (c^2 + c d + c^2) Reduce[fun, #] & /@ {A, B} // MatrixForm


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Here's my approach using replacement rule /. Sum[r^p_. a__, b_] :> r^p Sum[Times[a], b]. I have a hunch that there's a built in function that could accomplish this but I can't seem to find it. I've also made some modifications to your code: 1) Changing the first function to β to avoid clashing with the subscripted symbol α. 2) Changing the upper limit ...


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expr = aaa - 3 bbb + ccc + ddd - eee expr /. _?Negative :> 0 (* aaa + ccc + ddd *)


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Is something like this what you're after? Simplify[PiecewiseExpand[-Abs[m1 - m2] - Abs[m2 - m3] - Abs[m1 - m3] + n/2*(Abs[m1] + Abs[m2] + Abs[m3]), {m1 == m2, m2 >= m3}, Reals], {m1 == m2, m2 >= m3}] It seems nice and simple.


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As there are infinite solutions you must limit the scope of your search. You can find your desired a == b == c ==1 with f.e.: FindInstance[ a/(b + c) == b/(c + a) == c/(a + b) && 0 < a < 10 && 0 < b < 10 && 0 < c < 10, {a, b, c}, Integers, 9] {{a -> 1, b -> 1, c -> 1}, {a -> 2, b ...


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Mathematica can solve this problem completely. It provides three types of non-trivial solutions. Here we go The two defining eqations are eq1 = a/(b + c) == b/(c + a); eq2 = b/(c + a) == c/(a + b); Solving these, one by one, using Solve without specifying the variables to solve for Solve [eq1] {{a -> b}, {a -> -b - c}} Solve [eq2] {{a ...


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Perhaps what you looking for is Reduce. It will give the whole solution space of your equations. Reduce[a/(b + c) == b/(a + c) == c/(a + b), {a, b, c}, Reals] This should make it clear that a == b == c == 1 is not the answer. At the request of the OP in a comment below, here is a non-trivial solution where the three variables are not equal. ratios = ...



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