The Power of Absolute

Let $\begin{aligned} f(x) = 3\left(9+3^{|x|+|x+4|}\right) - 3^{|x+4|} - 3^{|x|+4} = 0 \end{aligned}$ , We have to consider $3$ cases when $f(x)$ changes form.

$\underline{\text{Case 1:}} \quad x > 0$

$\begin{aligned} 3\left(9+3^{x+x+4}\right) - 3^{x+4} - 3^{x+4} & = 0 \\ 3^3 + 3^5 (3^x)^2 - 2\dot{}3^4(3^x) & = 0 \\ 9(3^x)^2 - 6(3^x) + 1 & = 0 \\ \left(3(3^x) -1\right)^2 & = 0 \\ 3^x & = \frac{1}{3} \\ \Rightarrow x & = \color{#D61F06}{-1< 0 \quad \text{rejected}} \end{aligned}$

$\underline{\text{Case 2:}} \quad -4 < x \le 0$

$\begin{aligned} 3\left(9+3^{-x+x+4}\right) - 3^{x+4} - 3^{-x+4} & = 0 \\ 3^3 + 3^5 - 3^4(3^x) - 3^4(3^{-x}) & = 0 \\ 3(3^x)^2 - 10(3^x) + 3 & = 0 \\ (3(3^x) -1) ((3^x) -3) & = 0 \\ 3^x & = \begin{cases} \frac{1}{3} & \Rightarrow x = \color{#3D99F6}{-1 \in (-4, 0]} & \color{#3D99F6}{\text{accepted}} \\ 3 & \Rightarrow x = \color{#D61F06}{1 > 0} & \color{#D61F06} {\text{rejected}} \end{cases} \end{aligned}$

$\underline{\text{Case 2:}} \quad x \le -4$

$\begin{aligned} 3\left(9+3^{-x-x-4}\right) - 3^{-x-4} - 3^{-x+4} & = 0 \\ 3^3 + 3^{-3}(3^{-x})^2 - 3^{-4}(3^{-x}) - 3^4(3^{-x}) & = 0 \\ 3(3^{-x})^2 -6562(3^{-x}) + 2187 & = 0 \\ (3(3^{-x}) -1) ((3^{-x}) -2187) & = 0 \\ 3^{-x} & = \begin{cases} \frac{1}{3} & \Rightarrow x = \color{#D61F06}{1 > -4} & \color{#D61F06}{\text{rejected}} \\ 3^7 & \Rightarrow x = \color{#3D99F6}{-7 < -4} & \color{#3D99F6} {\text{accepted}} \end{cases} \end{aligned}$

Therefore the sum of solutions is $-1-7 = \boxed{-8}$

$3(9+{ 3 }^{ IxI\quad +\quad Ix+4I })\quad =\quad { 3 }^{ Ix+4I }+\quad { 3 }^{ IxI+4 }\quad \\ \Longrightarrow \quad 27\quad +\quad { 3 }^{ IxI + Ix+4I + 1 }=\quad { 3 }^{ Ix+4I }+{ 3 }^{ IxI +1 }.27\\ \Longrightarrow { 3 }^{ IxI+1 }({ 3 }^{ Ix+4I }-27)\quad =\quad { 3 }^{ Ix+4I }-27\\ now\quad there\quad are\quad two\quad ways\quad in\quad which\quad this\quad equation\quad would\quad satisfy\\ case1-\\ { 3 }^{ IxI+1 }=1\\ \Longrightarrow IxI\quad +\quad 1\quad =\quad 0\\ \Longrightarrow IxI=-1\\ which\quad is\quad not\quad possible\quad since\quad absolute\quad value\quad of\quad any\quad number\quad is\quad \\ always\quad positive.\quad so\quad this\quad case\quad is\quad not\quad possible.\\ case2-\\ { 3 }^{ Ix+4I }-27=0\\ \Longrightarrow { 3 }^{ Ix+4I }={ 3 }^{ 3 }\\ \Longrightarrow Ix+4I=3\\ which\quad yeilds\quad two\quad solutions\quad \\ x=-1,-7\\ since\quad no\quad other\quad case\quad is\quad possible.\quad these\quad will\quad be\quad the\quad only\quad two\quad solutions.\\ so\quad the\quad final\quad answer,\quad i.e\quad their\quad sum\quad would\quad be\quad -1\quad +\quad -7\quad =\quad -8\\ \\ \\ \quad$

$Let\quad a={ 3 }^{ \left| x \right| }\quad and\quad b={ 3 }^{ \left| x+4 \right| }\\ \Rightarrow 3(9+ab)=b+81a\\ \Rightarrow 27-b=3a(27-b)\\ \therefore a=\frac { 1 }{ 3 } \quad or\quad b=27\\ Now\quad a=\frac { 1 }{ 3 } \Rightarrow { 3 }^{ \left| x \right| }={ 3 }^{ -1 }\Rightarrow \left| x \right| =-1\quad which\quad is\quad impossible\quad as\quad \left| x \right| \ge 0\quad \forall x\\ Thus\quad b=27\Rightarrow { 3 }^{ \left| x+4 \right| }={ 3 }^{ 3 }\Rightarrow \left| x+4 \right| =3,\quad \therefore x+4=\pm 3\Rightarrow x=-7\quad or\quad -1\\ Thus\quad the\quad sum\quad of\quad all\quad real\quad solutions=\quad -1+-7=\boxed { -8 } \\ \\ \\$