Solving a inequality over the reals

Question

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 $0$.

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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?

Answer 1

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.

Answer 2

If you divide by -3d you first of all have to rule out $d=0$. [Edit: I just see, that $d,z\neq 0$]

Also you have to seperate cases, because if $-3d<0$ the 'direction' of the inequality changes.

It is $a\leq b\Leftrightarrow -a\geq -b$