Given x∈(0,1)x\in(0,1), I am interested in finding the solution of polynomials in the form of xa(1−x)b=c,x^a(1-x)^b=c, where a,ba,b are both positive integers, and cc is carefully chosen so that a root in (0,1)(0,1) exists.

For a+b<5a+b<5 the roots can be found painstakingly with radicals, however the problem won't be straightforward for larger a,ba,b due to the Abelâ€“Ruffini theorem. Generally, I know a root can be found using Jacobi's ϑ\vartheta functions, but their complexity makes them very difficult to apply in practice. So here is my question: can the root to this polynomial in (0,1)(0,1) be expressed, or approximated, by a "nicer" function? By "nicer", I mean a combination of elementary functions and "common" special functions (e.g. Beta, Bessel, etc.). ================= 1 @Nemo This equation can be reduced to trinomial equation How? Note that the a,ba,b here and the nn in the trinomial equation are all assumed to be positive integers. – dxiv yesterday ================= 1 Answer 1 ================= It's always hard to prove that a simple closed form does not exist, but I don't see one here. That said, for a,b>0a,b \gt 0 (not necessarily integers) f(x)=xa(1−x)bf(x)=x^a(1-x)^b is a nice concave function on [0,1][0,1], with f(0)=f(1)=0f(0)=f(1)=0 and a maximum at x=aa+bx=\frac{a}{a+b}.

It follows that for every 0≤c