How to find the one real root of $(x-1)(x-2)(x-3) + 1$, manually? The polynomial 
$$ (x-1)(x-2)(x-3) + 1 $$
has one root (I have seen it by a plot in LaTeX), $0 < x_{root} < 1$. So I presume that the polynomial can be rewritten as :
$$ (x-1)(x-2)(x-3) + 1 = (x - a)^{3} $$
but this is not possible, since
$$  (x-1)(x-2)(x-3)+1 = x^{3} - 6x^{2} + 11x -5   $$
$$ (x-a)^{3} = x^{3} - 3ax^{2} + 3a^{2}x -a^{3} $$

How to find the form of the polynomial without the remainder? (also
  manually, without numerical method)

Thanks before.
 A: We can solve $ (x-1)(x-2)(x-3) + 1 = 0 $ for $x \in \mathbb{R}$ by manipulating it in a quadratic equation with some convenient variable substitutions as follows:
Expand the polynomial
$$ x^3-6 x^2+11 x-5 = 0 $$
Substitute $ y = x - 2 $ to get rid of the quadratic term
$$ (y+2)^3 - 6(y+2)^2 + 11(y+2) - 5 = y^3 - y + 1 = 0 $$
Let $c \in \mathbb{R}$ be a constant we yet don't know its value but will determine a suitable value later. Substitute $ y = z + \frac{c}{z} $
$$ (z + \frac{c}{z})^3 - (z + \frac{c}{z}) + 1 = 0 $$
Multiply both sides by $z^3$ and group by $z$
$$ z^6 + z^4(3c-1) + z^3 + z^2c(3c-1)+c^3 = 0 $$
Note that by setting $ c = \frac{1}{3} $ we get rid of both the quadratic and the quartic terms yielding
$$ z^6 + z^3 + c^3 = 0 $$
To make computations easier make $ k = c^3$. Now substitute $ u = z^3 $ and we get a quadratic equation in $u$
$$ u^2 + u + k = 0 $$
Can you carry on from here? Let me know if you get stuck.
A: Short answer:
Factoring a polynomial in linear or quadratic factors and finding its roots are equivalent tasks, one solves the other.
Your polynomial has no simple form and you can't escape Cardano/Tartaglia's formulas, which are a little heavy, though tractable by hand https://en.wikipedia.org/wiki/Cubic_function#General_solution_to_the_cubic_equation_with_real_coefficients 
https://www.wolframalpha.com/input/?i=roots+of+(x-1)(x-2)(x-3)%2B1
