Change the length scale on the axes of original system of coordinates, in which the equation $$y=x^3-px\qquad\text{(1)}$$ is plotted, i.e. introduce new coordinates $x_1$ and $y_1$ instead of original $y$ and $x$, assuming $$x=kx_1, \quad y=ly_1, \qquad \text{(2)}$$ where $k\text{ and}\ y \in\mathbb{R^+_0}$, so that the equation $\text(1)$ after the transformation of coordinates $\text(2)$ would have the following look: $$y_1=x_1^3-x_1$$
I have found this problem in a textbook “Calculus for school students” with a mark “Easy to solve” and now I am stuck having done only this:
$$l(x_1^3-x_1)=k^3x_1^3-pkx_1$$ I failed expressing $k$ and $l$ in terms of $p$, as I guess it should be $p$-dependent. Otherwise, would the unit of length in new system of coordinates be changing?
Hint
Let us start with the original equation $$y=x^3-px$$ and apply $y=ly_1$ and $x=kx_1$ as required. Replacing then leads to $$ly_1=k^3x_1^3-pkx_1$$ that is to say $$y_1=\frac{k^3}{l} x_1^3- \frac{pk}{l}x_1$$ So, to get the required form, you must have $$\frac{k^3}{l}=1$$ and $$\frac{pk}{l}=1$$ So, two equations for two unknowns $k$ and $l$ the solutions of which depending on $p$.
I am sure that you can take from here.