If a line is linear
$$y= mx+b$$
If I cube the expression
$$y^{3} = (mx +b)^{3}$$
And now add a constant
$$y^{3} = (mx +b)^{3}+C$$
Roughly where is the line still linear if I take the cube root again?
$$y = ((mx +b)^{3}+C)^{1/3}$$
$$ y = (b^3+3 b^2 m x+3 b m^2 x^2+m^3 x^3 + C)^{1/3}$$
I plot it, I have something that looks very linear after a certain point
We have that
$$\begin{align*} y &= \sqrt[3]{(mx+b)^3+C} \\ &= \sqrt[3]{(mx+b)^3\cdot\left(1+\dfrac{C}{(mx+b)^3}\right)} \\ &= (mx+b)\cdot\sqrt[3]{1+\dfrac{C}{(mx+b)^3}} \\ \end{align*}$$
As $x\to\infty$, or as $x\to-\infty$, we have that $\dfrac{C}{(mx+b)^3}\to0$ and $\sqrt[3]{1+\dfrac{C}{(mx+b)^3}}\to1$.
So $\sqrt[3]{(mx+b)^3+C}$ will get closer to $mx+b$ if we make $x$ very big or very negative.
Edit: When I wrote the above, I was assuming that $m\ne0$. Of course, if $m=0$ and $y=\sqrt[3]{(mx+b)^3+C}$, then $y$ is constant.
