Why after depressing a cubic does it have different roots? The cubic $(x-1)(x-2)(x-2)=0$ will have the roots 1 and 2. Expanding will give $x^3-5x^2+8x-4=0$ which is in the form $ax^3+bx^2+cx+d=0$. Depressing it by substituting $x = t - \frac{b}{3a} = t+\frac{5}{3}$ will give $t^3-\frac{1}{3}t+\frac{2}{27}=0$. This is in the form $t^3+pt+q=0$.
Using Cardano's formula $x=\sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}}$ for the above, the following root is gotten.
\begin{align}
x &= \sqrt[3]{-\frac{\frac{2}{27}}{2} + \sqrt{\left(\frac{\frac{2}{27}}{2}\right)^2+\left(\frac{-\frac{1}{3}}{3}\right)^3}} + \sqrt[3]{-\frac{\frac{2}{27}}{2} - \sqrt{\left(\frac{\frac{2}{27}}{2}\right)^2+\left(\frac{-\frac{1}{3}}{3}\right)^3}}\\
&= \sqrt[3]{-\frac{1}{27} + \sqrt{\left(\frac{1}{27}\right)^2+\left(-\frac{1}{9}\right)^3}} + \dots\\
&= \sqrt[3]{-\frac{1}{27} + \sqrt{\frac{1}{729}+\left(-\frac{1}{729}\right)}} + \dots\\
&= \sqrt[3]{-\frac{1}{27} + \sqrt{0}} + \sqrt[3]{-\frac{1}{27} - \sqrt{0}}\\
&= \sqrt[3]{-\frac{1}{27}} + \sqrt[3]{-\frac{1}{27}}\\
&= -\frac{1}{3} + -\frac{1}{3}\\
&= -\frac{2}{3}\\
\end{align}
However, the initial cubic didn't have $-\frac{1}{6}$ as one of its roots. If the depressed cubic $t^3-\frac{1}{3}t+\frac{2}{27}=0$ is plotted, it's completely different to the original, the other root being $\frac{1}{3}$.
I thought that depressing a cubic was simply rewriting it and not changing it. And if it does change what the cubic is, then why is it said that Cardano's formula works for all cubics, shouldn't it be said that it only works for depressed cubics?

As you said in the first paragraph, you get the depressed cubic by substituting
$x = t - \frac{b}{3a} = t+\frac53$.
Conversely, that means $t = x - \frac53$ in your example.
So if $x = 1$ is a root of the original cubic, it had better be true that 
$t = x - \frac53 = -\frac23$ is a root of the depressed cubic.
And if $t = -\frac23$ is a root of the depressed cubic, then $x = t+\frac53 = 1$
is a root of the original cubic.
And as it turns out that's exactly what happened.
