(ITA) Radical Power

$8^{\sqrt{x+1}} - 19 \times 4^{\sqrt{x+1}} + 44 \times 2^{\sqrt{x+1}} + 64 = 0$

Substituting $2^{\sqrt{x+1}} = a$ ,

$a^3- 19a^2 + 44a + 64 = 0$

$a^3 + a^2 -20a^2 - 20a + 64a + 64 = 0$

$a^2(a+1)-20a(a+1)+64(a+1) = 0$

$(a+1)(a^2-20a+64) = 0$

$(a+1)(a-16)(a-4)=0$

$(2^{\sqrt{x+1}}+1)(2^{\sqrt{x+1}}-16)(2^{\sqrt{x+1}}-4)=0$

Now, $2^{\sqrt{x+1}} - 16 = 0 \longrightarrow x = 15$

$2^{\sqrt{x+1}} - 4 = 0 \longrightarrow x = 3$

$2^{\sqrt{x+1}} +1 = 0$ has no real value.

Hence, the answer is $15+3 = \boxed{18}$

$8^{\sqrt{x+1}} - 19 \cdot 4^{\sqrt{x+1}} + 44 \cdot 2^{\sqrt{x+1}} + 64 = 0$ $y = 2^{\sqrt{x+1}} \Rightarrow y^3 - 19y^2 + 44y + 64 = 0$ $1 + 44 = -19 + 64 \Rightarrow y = -1$ $(y+1)(y^2-20y+64)=0$ $V_y = \left \{ -1, 4 , 16 \right \}$ $2^{\sqrt{x+1}}= -1, \; 2^2, \; 2^4$ $2^k \geq 0, \; \forall \; k \in \mathbb{R}$ $\sqrt{x+1} = 2, \; \sqrt{x+1} = 4$ $V_x = \left \{ 3, 15 \right \}$ $\boxed{\Sigma(x) = 18.}$