So i started with the equation $$z^3+az^2+bz+c=0,$$ with $z\in\mathbb{C}$. I've set $z=w+h$ and chose $h=-\frac{1}{3}a$ to get the equation \begin{align} w^3+Bw+C=0, \end{align} with \begin{align*} B&= -\frac{1}{3}a^2+b \\ C&= \frac{2}{27}a^3-\frac{1}{3}ab+c. \end{align*}
Now I'm asked to take $w=kv$ and to manipulate $w^3+Bw+C=0$ into $$4v^3-3v+A=0.$$
First I thought to take $k^3=4$ and hence $k=\sqrt[3]{4}$, so $B$ should have been equal to $-\frac{3}{\sqrt[3]{4}}$. But then I found out I already calculated $B$ so I cannot 'choose' it.
What is a better way to do this?
Hint: I think you're on the right track. If you have $$w^3+Bw+C=0$$ then use $$w=kv$$ and use a constant scaling factor $r$ on both sides to get $$rk^3v^3 +Brkv +rC=0$$Now you just need to find $k$ and $r$ satisfying $$\begin{cases} rk^3 = 4\\Brk = -3 \end{cases}$$Then your $A$ is $rC$.