A hypergeometric beast: Simple closed-form for combination of $_4F_3(1)$? We have
$$
\frac{1}{4}
-\frac{1}{4}{_4F}_3\left({-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5}\atop\frac{1}{4},\frac{1}{2},\frac{3}{4}};1\right)
-\frac{1}{\sqrt[4]
   {5}}{_4F}_3\left({\frac{1}{20},\frac{9}{20},\frac{13}{20},\frac{17}{20}\atop\frac{1}{2},\frac{3}{4},\frac{5}{4}};1\right)\\
+\frac{1}{5\sqrt{5}}{_4F}_3\left({\frac{3}{10},\frac{7}{10},\frac{9}{10},\frac{11}{10}\atop\frac{3}{4},\frac{5}{4},\frac{3}{2}};1\right)
-\frac{7}{50\,5^{3/4}}{_4F}_3\left({\frac{11}{20},\frac{19}{20},\frac{23}{20},\frac{27}{20}\atop\frac{5}{4},\frac{3}{2},\frac{7}{4}};1\right)\\
=-\frac{1}{4}\left(1+\sqrt[3]{5/9}\left(\sqrt[3]{9+4\sqrt 6}-\sqrt[3]{4\sqrt 6-9}\right)\right).
$$
Through indirect methods, this relationship has been proven true by showing that both sides of the equality are solutions to
$$
4x^3+3x^2+2x+1=0.
$$

How can we prove this relation directly by reducing the hypergeometric functions?

While I have some experience with reducing hypergeometric functions, I don't even know where to begin with this one due to the rational parameters with denominator $20$. My initial thought were to work with the integral representation DLMF 16.5.2 to write the $_4F_3(1)$ functions as double integrals of $_2F_1(\cdot)$. For example, taking the first hypergeometric function above we have
$$
{_4F}_3\left({-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5}\atop\frac{1}{4},\frac{1}{2},\frac{3}{4}};1\right)=(constant)\int_0^1\int_0^1u^{\frac{4}{20}-1}(1-u)^{\frac{1}{20}-1}v^{\frac{8}{20}-1}(1-v)^{\frac{2}{20}-1} {_2F_1}\left({-\frac{4}{20},\frac{12}{20}\atop \frac{15}{20}};uv\right)\,\mathrm du\mathrm dv.
$$
At least in this form there seems to be some hope of reducing the $_2F_1(\cdot)$ function in the integrand, which may then lead to the desired form.

Remark
There is this paper
Beukers, Frits, Hypergeometric functions, how special are they?, Notices Am. Math. Soc. 61, No. 1, 48-56 (2014). ZBL1323.33007.
Plugging into Theorem 2 of that paper, we find that
$$
{_4F}_3\left({-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5}\atop\frac{1}{4},\frac{1}{2},\frac{3}{4}};x\right),\quad
{_4F}_3\left({\frac{1}{20},\frac{9}{20},\frac{13}{20},\frac{17}{20}\atop\frac{1}{2},\frac{3}{4},\frac{5}{4}};x\right),\quad
{_4F}_3\left({\frac{3}{10},\frac{7}{10},\frac{9}{10},\frac{11}{10}\atop\frac{3}{4},\frac{5}{4},\frac{3}{2}};x\right),\quad
{_4F}_3\left({\frac{11}{20},\frac{19}{20},\frac{23}{20},\frac{27}{20}\atop\frac{5}{4},\frac{3}{2},\frac{7}{4}};x\right)
$$
are all algebraic functions.  Probably you would have to consult the references of that paper to find explicit polynomials that they satisfy.

Maple code
> # Beukers, Notices, January 2014, p. 48

> reducemod1:=proc(s)
> {seq(x-floor(x),x in s)};
> end;
> 
> isirreducible := proc(l1,l2)
> local s1,s2;
> s1:=reducemod1({op(l1)});
> s2:=reducemod1({op(l2)} union {1});
> if s1 intersect s2 = {} then return(true) else return(false) fi;
> end;
> 
> isinterlaced := proc(l1,l2)
> isinterlacedL(reducemod1({op(l2)} union {1}), reducemod1({op(l1)}));
> end;
> 
> isinterlacedL := proc(l1,l2)
> local s1,x;
> x:=min(l1);
> if x >= min(l2) then return(false); fi;
> s1 := l1 minus {x};
> if s1 = {} then 
>  if nops(l2) <= 1 then return(true) else return(false); fi; 
> fi;
> isinterlacedL(l2,s1);
> end;
> 
> thm2 := proc(l1,l2) # is it algebraic?
> local s1,s2,D,r,x;
> s1:= reducemod1(l1);
> s2:= reducemod1(l2) union {0};
> if s1 intersect s2 <> {} then return("reducible"); fi;
> D:=lcm(seq(map(denom,s1 union s2)));
> for r from 1 to D-1 do
> if igcd(r,D) = 1 then
>   if not isinterlaced({seq(r*x,x in s1)},{seq(r*x,x in s2)}) then return("transcendental"); fi;
> fi;
> od;
> return("algebraic");
> end;
>
> thm2([1/4,1/4],[1/2]);
                        "transcendental"

> thm2([3/10,7/10,9/10,11/10],[3/4,5/4,3/2]);
                          "algebraic"

