The sum of all the roots of the equation $$(x-1)^2-5\mid x-1\mid+\ 6=0$$ is:

We have $$(x-1)^2- 5|x-1| +6=0$$

Expanding and simplifying, we get,

$$x^2+1-2x - 5|x-1| + 6 = 0$$

$$\Rightarrow x^2 - 2x - 5|x-1|+7 = 0$$

Case A: $$x\geq 1$$

The equation simplifies to

$$x^2-2x-5x+5+7=0$$

$$\Rightarrow x^2 - 7x + 12 = 0$$

The sum of roots of the quadratic equation above is $$\dfrac{-b}{a} = \dfrac{-(-7)}{1} = 7$$

Case B: $$x<1$$

The equation simplifies to

$$x^2-2x+5x-5+7 = 0$$

$$\Rightarrow x^2+3x+2=0$$

The sum of roots of the quadratic equation above is $$\dfrac{-b}{a} = \dfrac{-3}{1} = -3$$

Thus, the sum of all roots of the given equation is $$7+(-3) = 4$$. 

Option A is the correct answer.