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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?

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  • $\begingroup$ That's what I get for wiring this at half past midnight. It still gives a negative sixth though so point still stands; let me fix the post. $\endgroup$
    – ettolrach
    Mar 8, 2020 at 0:29
  • $\begingroup$ I wanna say I did that because I was thinking of 0.6 but I might just be making excuses for myself :P. The roots should definitely be -2/3 and 1/3. $\endgroup$
    – ettolrach
    Mar 8, 2020 at 0:39
  • $\begingroup$ Right, now that is $t=-2/3$, and $x=t+5/3=1$ (similarly the other one). This substitution changed the roots. $\endgroup$
    – Sil
    Mar 8, 2020 at 0:42
  • $\begingroup$ Ah, so to get the roots of the original I'd need to add on $\frac{5}{3}$ again? That would give me $1$ and $2$ as expected. $\endgroup$
    – ettolrach
    Mar 8, 2020 at 0:45
  • $\begingroup$ Yes, that's what that substitution means $\endgroup$
    – Sil
    Mar 8, 2020 at 0:48

1 Answer 1

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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.

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  • $\begingroup$ After all, if two monic cubics don’t look the same, they have different root-sets. $\endgroup$
    – Lubin
    Mar 8, 2020 at 2:47
  • $\begingroup$ @Lubin True, all you can do really (without changing roots) is multiply all coefficients by a non-zero constant. $\endgroup$
    – David K
    Mar 8, 2020 at 4:40
  • $\begingroup$ And then you get a nonmonic polynomial. $\endgroup$
    – Lubin
    Mar 8, 2020 at 17:56

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