Complex 360!

The substitution $(x+3)^2=t$ leaves us with $t(t-1)(t-4)-360=(t-9)(t^2+4t+40)=0$ so that $t=-2\pm 6i$ and $x=\pm\sqrt{\sqrt{10}-1}-3\pm\sqrt{\sqrt{10}+1}i$ . The answer is $\boxed{14}$

$\left( x+1 \right) \left( x+2 \right) { \left( x+3 \right) }^{ 2 }\left( x+4 \right) \left( x+5 \right) -360=0\\ \left( x+1 \right) \left( x+5 \right) { \left( x+3 \right) }^{ 2 }\left( x+2 \right) \left( x+4 \right) -360=0\\ \left( { x }^{ 2 }+6x+5 \right) \left( { x }^{ 2 }+6x+9 \right) \left( { x }^{ 2 }+6x+8 \right) -360=0\\$

Let $z={ x }^{ 2 }+6x$ .

$\left( z+5 \right) \left( z+8 \right) \left( z+9 \right) -360=0\\ z\left( { z }^{ 2 }+22z+157 \right) =0\\ \\ z=0\rightarrow { x }^{ 2 }+6x=0\rightarrow x=0,-6\quad (not\quad needed)\\ \\ { z }^{ 2 }+22z+157=0\rightarrow { z }^{ 2 }+22z+121=-36\\ z=-11\pm 6i\\ { x }^{ 2 }+6x=-11\pm 6i\\ { x }^{ 2 }+6x+9=-2\pm 6i$

{ \left( a+bi \right) }^{ 2 }=-2\pm 6i\\ a^{ 2 }+2abi-b^{ 2 }=-2\pm 6i\\ \\ \left\{ \begin{matrix} a^{ 2 }-b^{ 2 }=-2 \\ 2ab=\pm 6 \end{matrix} \right \\ \\ a^{ 2 }-\left( \pm \frac { 3 }{ a } \right) ^{ 2 }=-2\rightarrow a^{ 4 }-9=-2{ a }^{ 2 }\\ \left( \pm \frac { 3 }{ b } \right) ^{ 2 }-b^{ 2 }=-2\rightarrow 9-b^{ 4 }=-2{ b }^{ 2 }\\ \\ a^{ 4 }+2{ a }^{ 2 }+1=10\\ b^{ 4 }-2{ b }^{ 2 }+1=10\\ \\ { a }^{ 2 }+1=\pm \sqrt { 10 } \rightarrow a=\pm \sqrt { \sqrt { 10 } -1 } \\ { b }^{ 2 }-1=\pm \sqrt { 10 } \rightarrow b=\pm \sqrt { \sqrt { 10 } +1 } \\ \\ x+3=\pm \sqrt { \sqrt { 10 } -1 } \pm \sqrt { \sqrt { 10 } +1 } i\\ x=\pm \sqrt { \sqrt { 10 } -1 } -3\pm \sqrt { \sqrt { 10 } +1 } i

Therefore $x+y+z=10+1+3=14$ .