Prove a hypergeometric Identity: $\frac{\Gamma(\frac14)^2 }{8\sqrt{\pi}}\,_3F_2(.) +\frac1{2\pi}\,_3F_2(.) =\frac{\Gamma( \frac14 )^6}{32\pi^{7/2}}$

Question

Are there any direct ways to prove the following relation between two $\,_3F_2$s? $$ \frac{\Gamma\left ( \frac14 \right )^2 }{8\sqrt{\pi} } \,_3F_2\left ( \frac14,\frac14,\frac12;1,1;1 \right ) +\frac{1}{2\pi}\,_3F_2\left ( \frac34,\frac34,1;\frac32,\frac32;1 \right ) =\frac{\Gamma\left ( \frac14 \right )^6}{32\pi^{7/2}}. $$ The non-direct proof is based on observing the two equalities: \begin{aligned} &\,_4F_3\left ( \frac14,\frac12,\frac12,\frac12;1,1,\frac54;-1 \right ) = \frac{\Gamma\left ( \frac14 \right )^2 }{8\sqrt{\pi} } \,_3F_2\left ( \frac14,\frac14,\frac12;1,1;1 \right ), \\ &\,_4F_3\left ( \frac14,\frac12,\frac12,\frac12;1,1,\frac54;-1 \right )=-\frac{1}{2\pi}\,_3F_2\left ( \frac34,\frac34,1;\frac32,\frac32;1 \right ) +\frac{\Gamma\left ( \frac14 \right )^6}{32\pi^{7/2}}, \end{aligned} where the second one refers to arXiv:1806.08411(p16). I would appreciate it if you provide me with some ideas.