<s>[INST] 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.
 [/INST] 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"

</s>