Sum them

$\begin{aligned} \sqrt[4]{1-x} + \sqrt[4]{15+x} & = 2 & \small \color{#3D99F6} \text{Let }y=x+7 \\ \sqrt[4]{8-y} + \sqrt[4]{8+y} & = 2 & \small \color{#3D99F6} \text{Raise both sides to the power of }4 \\ 8 - y + 4(8-y)^\frac 34 (8+y)^\frac 14 + 6(8-y)^\frac 12 (8+y)^\frac 12 + 4(8-y)^\frac 14 (8+y)^\frac 34 + 8 + y & = 16 \\ 4(8-y)^\frac 12 (64-y^2)^\frac 14 + 6(64-y^2)^\frac 12 + 4(64-y^2)^\frac 14 (8+y)^\frac 12 & = 0 \\ (64-y^2)^\frac 14\left(2(8-y)^\frac 12 + 3(64-y^2)^\frac 14 + 2(8+y)^\frac 12\right) & = 0 \\ (64-y^2)^\frac 14\left(2(8-y)^\frac 12 + 4(8-y)^\frac 14 (8+y)^\frac 14 + 2(8+y)^\frac 12 - (8-y)^\frac 14 (8+y)^\frac 14 \right) & = 0 \\ (64-y^2)^\frac 14\left(2 {\color{#3D99F6}\left(\sqrt[4]{8-y}+ \sqrt[4]{8-y}\right)^2}- (8-y)^\frac 14 (8+y)^\frac 14 \right) & = 0 & \small \color{#3D99F6} \text{Note that }\sqrt[4]{8-y}+ \sqrt[4]{8-y} = 2 \\ (8-y)^\frac 14 (8+y)^\frac 14 \left(8- (64-y^2)^\frac 14 \right) & = 0 \end{aligned}$

$\implies y = \begin{cases} -8 & \implies x = -15 & \color{#3D99F6} \text{A solution} \\ 0 & \implies x = - 7 & \color{#D61F06} \text{Not a solution} \\ 8 & \implies x = 1 & \color{#3D99F6} \text{A solution} \end{cases}$

Therefore, the sum of solutions is $-15+1=\boxed{-14}$ .

${\mathrm{(}\sqrt[4]{1-x}\mathrm{\ +}\sqrt[4]{15+x}\mathrm{)}}^4 = 2^4$ $16 + 4\sqrt[4]{{\left(1-x\right)}^3(15+x)} + 6\sqrt[4]{{\left(1-x\right)}^2{\left(15+x\right)}^2} + \sqrt[4]{\left(1-x\right){\left(15+x\right)}^3} = 16$ $4\sqrt[4]{{\left(1-x\right)}^3(15+x)} + 6\sqrt[4]{{\left(1-x\right)}^2{\left(15+x\right)}^2} + \sqrt[4]{\left(1-x\right){\left(15+x\right)}^3} = 0$ $\sqrt[4]{1-x}\sqrt[4]{15+x}(4\sqrt[4]{{\left(1-x\right)}^2} + 6\sqrt[4]{{\left(1-x\right)}{\left(15+x\right)}}+\sqrt[4]{{\left(15+x\right)}^2})= 0$ $4\sqrt[4]{{\left(1-x\right)}^2} + 6\sqrt[4]{{\left(1-x\right)}{\left(15+x\right)}}+ \sqrt[4]{{\left(15+x\right)}^2}$ is increasing so x= 1 V x= -15 , the sum is -14