Solving a inequality over the reals I have the following inequality that I need to solve for real $z$:
$$3\cdot\left(-d\left(d^3+4z^3-168\right)\right)\ge0\tag1$$
Where $d$ and $z$ are element of the real numbers (so they can be positive and negative) and they are not equal to $0$.

My work:
I first divided both sides by $-3d$, and it got me:
$$d^3+4z^3-168\ge0\tag2$$
So, I can also write:
$$z^3\ge\frac{8-d^3}{4}\tag3$$
And can I now just write:
$$z\ge\left(\frac{8-d^3}{4}\right)^\frac{1}{3}\tag4$$

Is my formula $(4)$ right?


You have to be careful when dividing with $-3d$, because as you said it can be either positive or negative. For some reason your last formula has $8$ instead of $168$ in it. But even after you correct that mistake, your formula $(4)$ is only valid for $d<0$, and for $d>0$ you need to flip the inequality sign.
