Interaction potential in standard $phi^4$ theory

In this paper, the authors consider a real scalar field theory in $d$-dimensional flat Minkowski space-time, with the action given by $$S=\int d^d\! x \left[\frac12(\partial_\mu\phi)^2-U(\phi)\right],$$ where $U(x)$ is a general self-interaction potential. Then, the authors proceed by saying that for the standard $\phi^4$ theory, the interaction potential can be written as $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2.$$

Why is this so? What is the significance of the cubic term present?

EDIT: Comparing with the scalar field theory the potential term involved with $\lambda$. What is the value we have inserted here?

Moreover I want to transform the potential to a new form $$U(\phi)= \frac{1}{8} \lambda(\phi ^2 -v^2)^2.$$ (I have got this from Mark Srednicki page no 576. )

Answer

Edit: Oh, I see that this is basically what twistor59 wrote in a comment.

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Set $\psi = \phi - 1$. The Lagrangian written in terms of $\psi$ is

$$ S = \int |\nabla \psi|^2 + V(\psi) $$

where $V(\psi) = U(\phi) = U(\psi + 1)$, expanding the reparametrization we have

$$ V(\psi) = \frac18 (\psi + 1)^2 (\psi - 1)^2 = \frac18 (\psi^4 - 2 \psi^2 + 1) $$

The constant 1 is not dynamical so the potential is equivalent to

$$ \tilde{V}(\psi) = \frac18 (\psi^4 - 2 \psi^2) $$

which is equivalent to the standard $\psi^4$ potential without cubic term up to scaling constants.

That is to say: the $\phi^3$ term isn't really there: it can be gauged away.