The sum of all real roots of the equation

|x − 1|² − 5|x − 1| + 6 = 0

We are given the equation:
${|x-1|}^2-5|x-1|+6=0$

To solve this equation, we can use a substitution method. Let $y=|x-1|$.
Since $y$ represents an absolute value, we must have the condition $y\ge 0$.

Substituting $y$ into the original equation gives a quadratic equation in terms of $y$:
$y^2-5y+6=0$

We can factor this quadratic equation by finding two numbers that multiply to 6 and add to -5. These numbers are -2 and -3:
$(y-2)(y-3)=0$

This gives two possible values for $y$:
$y=2$ or $y=3$

Both of these values satisfy the condition $y\ge 0$. Now we substitute back $y=|x-1|$ to find the real roots for $x$.

Case 1: $|x-1|=2$
This absolute value equation yields two equations:
$x-1=2\Rightarrow x=3$
$x-1=-2\Rightarrow x=-1$

Case 2: $|x-1|=3$
This absolute value equation also yields two equations:
$x-1=3\Rightarrow x=4$
$x-1=-3\Rightarrow x=-2$

Thus, the set of all real roots of the equation is $\{-2,-1,3,4\}$.

We are asked to find the sum of all real roots:
$\text{Sum of roots}=(-2)+(-1)+3+4$
$\text{Sum of roots}=-3+7=4$

Therefore, the sum of all real roots of the equation is 4.