Knowing binomial expansion of $(1+kx)^n$ shortcut for $(a+kx)^n$ If I've calculated the binomial expansion of $(1+kx)^n$ (where $k$ and $n$ are known) up to some term, is there then a shortcut for calculating $(a+kx)^n$?

Example: Given that the binomial expansion of $(1+6x)^{\frac23}$ up to and including the term in $x^2$ is $1+4x-4x^2$, what is the binomial expansion of $(8+6x)^{\frac23}$ (up to and including the term in $x^2$)?


Hint: Let's assume you already know $$f(x)=(1+kx)^n$$
Since 
  \begin{align*}
(a+kx)^n=a^n\left(1+k\frac{x}{a}\right)^n\tag{1}
\end{align*}
   you can calculate
\begin{align*}
(a+kx)^n=a^nf\left(\frac{x}{a}\right)\tag{2}
\end{align*}

Comment: Some details due to OPs comment


*

*In (1) we use
\begin{align*}
(a+kx)^n&=\left(\frac{a}{a}\left(a+kx\right)\right)^n\\
&=\left(a\cdot\frac{a+kx}{a}\right)^n\\
&=\left(a\left(1+\frac{kx}{a}\right)\right)^n\\
&=a^n\left(1+k\frac{x}{a}\right)^n\tag{3}\\
\end{align*}

*In (2) we use
\begin{align*}
a^nf\left(\frac{x}{a}\right)=a^n\left(1+k\frac{x}{a}\right)^n=(a+kx)^n
\end{align*}
according to (3)
