Two Two Three Three Four Four

$(x+2)^2+(x+3)^3 =(x+4)^4$

$\Rightarrow (x^2 + 4x + 4) + (x^3 + 9x^2 + 27x + 27)$ $\quad \space = x^4 + 16x^3 + 96x^2 + 256x + 256$

$\Rightarrow x^4 + 15x^3 + 86x^2 + 225x + 225 = 0$

$\Rightarrow (x+3)(x+5)(x^2 + 7x + 15) = 0$

$\Rightarrow x = -3$ , $x=-5$ and

$\quad \quad \quad \quad x = \dfrac {-7 \pm \sqrt{7^2-4(2)(15)}}{2} = \dfrac {-7 \pm \sqrt{-11}}{2} = \dfrac {-7 \pm \sqrt{11}i}{2}$

$\Rightarrow a= 7$ , $b=11$ and $c=2\quad \Rightarrow a+b+c = 7+11+2 = \boxed {20}$

\text{To simplify calculations, y=x+4 so f(y)= (y-2)^2+(y-1)^3 -y^4}\\f(1)=0, f(-1)=0,\therefore \dfrac{f(y)}{y^2-1}=0\text{ gives function with complex roots.}\\\therefore ~ \dfrac{f(y)}{y^2-1}=y^2-y+3=0.~~\implies~y=\dfrac{1\pm \sqrt{1^2-12}}{2}=\dfrac{1\pm \sqrt{11}i}{2}\\=x+4.~\therefore~a=7,b=11,c=2

After calculating we get xxxx+15xxx+86xx+225x+225=0,This Polynomial can be written as (xx+7x+15)(xx+8x+15)=0.so complex roots are (-7±i√11)/2 ,so a=7,b=11,c=2 and a+b+c=20

For an even slightly easier calculation, we can set $z=x+3$ , so the equation becomes $(z-1)^2+z^3=(z+1)^4.$ Then z=0 and z=-2 are two fairly self-evident real roots, and the variable change gives us x=-3 and x=-5, from which we can use synthetic division to find that the remaining quadratic polynomial is $x^2+7x+15.$ The solution follows.