Is solubility coefficient affected if ion data is given in Ksp? Recently I've been looking at this website, which says:

The $K_{\mathrm{sp}}$ of $\ce{Cr(OH)3}$ is $6.70 \times 10^{-31}$
Problem 1: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate?
$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$
$$K_{\mathrm{\mathrm{sp}}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$
$$[\ce{Cr^{3+}}]=s$$
$$[\ce{OH^{-}}]=3s$$
Problem 2: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate if the solution has $[\ce{Cr^{3+}}] = 0.0670~\mathrm{M}$?
$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$
$$K_{\mathrm{sp}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$
$$6.70 \times 10^{-31}=[0.0670~\mathrm{M}][s^{3}]$$

I don't understand why the site wrote this:

Note that $3s$ is not necessary. In example one, $s$ was assigned to the chromium ion and we knew the hydroxide to be three times greater than that value. Here, the hydroxide is simply an unknown value and it is not expressed in terms of some other unknown value (as it was in Problem 1).

Can anyone tell me why?
Also is also used to other equilibrium constant equation? What did I miss?
Let the concentration of $\ce{Cr^{3+}}$ be $t$ then the concentration of $\ce{OH-}$ will be $3t$.

So, the site says that writing the concentration as $3t$ is not necessary.

What it is basically doing is considering $3t$ equal to some variable $s$, and solving the question. There is nothing wrong in this, is it?

Now, for example we had not done that and calculated the value of $t$ by using the equation

$K_{\mathrm{sp}}=[\ce{Cr^{3+}}] \cdot (3t)^3$

You will then multiply the derived value of $t$ with 3 to get $\ce{[OH-]}$.

From which you will find the $\mathrm{pH}$. So the site says:

"Why not calculate the value of $3t$ directly by using a variable $s$?"

Therefore: $s=3t$

Therefore: $\mathrm{pOH}=-\log{s}$

What you are thinking is also correct, and the answer with that will be:
$\mathrm{pOH}=-\log{3t}$ 

So you both are doing the exact same thing here.