Deriving General Solution for Depressed Cubics

Question

I was reading An Imaginary Tale: The Story of the Square Root of Negative One and it begins with a derivation of the general solution for depressed cubics.

It begins with the format for depressed cubics:

$$x^3+px=q$$

Then says that $x$ can be written where $x=u+v$, which inserted into the original question gives

$$u^3+v^3+(3uv+p)(u+v)=q$$

It then says, "This single, rather complicated-looking equation, can be rewritten as two individually less complicated statements:"

$$3uv+p=0$$

and $$u^3+v^3=q$$

My question is why are these two equations the same as the one equation above?

score 0 · Answer 1 · answered Dec 22 '15 at 17:20

0

$$u^3+v^3+(3uv+p)(u+v)=q$$$$u^3+v^3=(u+v)(u^2-uv+v^2)$$$$(u+v)(u^2-uv+v^2)+(3uv+p)(u+v)=q$$$$(u+v)(u^2-uv+v^2+3uv+p)=(u+v)(u^2+v^2+2uv+p)$$$$u^2+2uv+(v^2+p)=(u-\frac{-2v+\sqrt{-4p}}{2})(u-\frac{-2v-\sqrt{-4p}}{2})$$$$(u+v)(u-\frac{-2v+\sqrt{-4p}}{2})(u-\frac{-2v-\sqrt{-4p}}{2})=q$$

Does that help?

answered Dec 22 '15 at 17:20

Simply Beautiful Art

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This doesn't answer the question. – Wojowu Dec 23 '15 at 14:16
@Wojowu I wasn't sure if it did to be honest. But I'd at least try – Simply Beautiful Art Dec 23 '15 at 14:27
@Wojowu Have you been stalking me on this site? I seem to be running into you a lot XD – Simply Beautiful Art Dec 23 '15 at 14:28
I have looked at your profile and saw you had some recent activity, so I decided to comment on it. – Wojowu Dec 23 '15 at 14:29
@Wojowu Oh, thanks. ;) Did you read my profile description? – Simply Beautiful Art Dec 23 '15 at 14:29

score 0 · Accepted Answer · answered Jan 28 '18 at 00:25

Start with the identity

$$(u+v)^3 -3uv(u+v) -(u^3+v^3) = 0$$

Rewrite your depressed cubic as $$x^3 +px-q=0$$ Substitute in $x =u+v$ as a root to get $$(u+v)^3+p(u+v) -q=0$$ Then note the correspondence between that equation and the initial identity, so $$q = (u^3+v^3)$$ and $$p = -3uv$$ and hence $$3uv+p=0$$

Deriving General Solution for Depressed Cubics

2 Answers2