Power transformation on interactions in linear regression In a multiple linear regression I have one predictor showing a cubic relationship with the outcome. If the same predictor $A$ is also part of an interaction with a continuous variable $B$, do I have to apply the polynomial transformation to both terms?
$$
\hat{Y} = {c} + \beta_1{A} + \beta_2{A^2} + \beta_3{A^3} + \beta_4{A}{B} + \beta_5{A^2}{B} +\beta_6{A^3}{B}.
$$
Logically it makes sense, but I've never seen a transformation applied to an interaction. Also I was wondering if the relationship can be simplified.
I've found some information in this thread but my question is simply to understand if the transformation is correct.
 A: I'm assuming you mean to have different coefficients for each term (they are all represented by $\beta$). 
Think of regression as attempting to fit the function $f$ such that 
$$ y = f(x_{1}, x_{2}, ..., x_{p}) + \varepsilon $$ 
The linear model for $f$ including only linear terms, 
$$ f(x_{1}, x_{2}, ..., x_{p}) = \beta_{0} + \beta_{1}x_{1} + ... + \beta_{p} x_{p}, $$ 
is the first order Taylor approximation of $f$, where the coefficients can be expressed in terms of the derivatives of $f$. If one additionally includes the squared terms (and cross products): 
$$\sum_{k} \alpha_{k} x_{k}^{2} + \sum_{k\neq j} \zeta_{kj} x_{k} x_{j} $$
then you have a second order taylor approximation of $f$, and similarly with higher order approximations. 
The model you have there is reflective of an $f$ where the fourth order taylor approximation is such that all coefficients on the terms including $B^2$ are 0. This is certainly possible and only means that the higher order derivatives of $f$ involving $B$ are 0. 
