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16 / 1 / 13 / ?

File: IMG_20220728_170542.jpg ( 2.89 MB , 2490x3106 , 1659038887951.jpg )

Image
  No. 257
>>258 >>452 >>335
no eureka for today, gentlemen
>>
  No. 258
>>257
You are trying to derive cubic formula?
>>
  No. 259 >>628
trying to use the same idea of ​​bisquare equations for a cubic equation...
>>
  No. 335
>>257
Keep trying...sooner or later, you'll understand.
>>
  No. 337
why not just use sage math? super ez to solve long equations
>>
  No. 452
>>257
We'll get em next time buddy
>>
  No. 628
>>259
Do you know Tschirnhaus transformation?? If yes you do, you'll can use it.
>>
  No. 631 >>639 >>633
Not OP, but I derived it 2 days ago while using toilet.

And it was for the most basic version of the problem:

x3+x+a=0 x^3 + x + a = 0


But it can probably be transformed into any cubic formula if you set
x=sy+tx = sy+t
and multiply left side by some constant.

I didn't really derive it, I just guessed it should be in form:


x=p+q3+p−q3 x = \sqrt[3]{p + q} + \sqrt[3]{p - q}


When I use it in the equation, I get:

(p+q3+p−q3)3+p+q3+p−q3+a=0 (\sqrt[3]{p + q} + \sqrt[3]{p - q})^3 + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0


p+q+3p+q3p+q3p−q3+3p+q3p−q3p−q3+p−q p + q + 3\sqrt[3]{p + q}\sqrt[3]{p + q}\sqrt[3]{p - q} + 3\sqrt[3]{p + q}\sqrt[3]{p - q}\sqrt[3]{p - q} + p - q

+p+q3+p−q3+a=0 + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0


2p+3p+q3p2−q23+3p−q3p2−q23+p+q3+p−q3+a=0 2p + 3\sqrt[3]{p + q}\sqrt[3]{p^2 - q^2} + 3\sqrt[3]{p - q}\sqrt[3]{p^2 - q^2} + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0


2p+3(p+q3+p−q3)p2−q23+p+q3+p−q3+a=0 2p + 3(\sqrt[3]{p + q} + \sqrt[3]{p - q})\sqrt[3]{p^2 - q^2} + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0


(2p+a)+(1+3p2−q23)(p+q3+p−q3)=0 (2p + a) + ( 1 + 3\sqrt[3]{p^2 - q^2})(\sqrt[3]{p + q} + \sqrt[3]{p - q}) = 0


In order for this equation to be satisfied I'll break this equation in two:

{2p+a=0(1+3p2−q23)(p+q3+p−q3)=0 \begin{cases} 2p + a = 0 \\ ( 1 + 3\sqrt[3]{p^2 - q^2})(\sqrt[3]{p + q} + \sqrt[3]{p - q}) = 0 \end{cases}


{p=−a21+3p2−q23=0 \begin{cases} p = -\frac{a}{2} \\ 1 + 3\sqrt[3]{p^2 - q^2} = 0 \end{cases}


Then we solve
qq
:

1+3−a24−q23=01 + 3\sqrt[3]{-\frac{a^2}{4} - q^2} = 0


−a24−q23=−13\sqrt[3]{-\frac{a^2}{4} - q^2} = -\frac{1}{3}


−a24−q2=−127-\frac{a^2}{4} - q^2 = -\frac{1}{27}


q2=−a24+127q^2 = -\frac{a^2}{4} + \frac{1}{27}


q=−a24+127q = \sqrt{-\frac{a^2}{4} + \frac{1}{27}}


and thus:

x=−a2+−a24+1273+−a2−−a24+1273 x = \sqrt[3]{-\frac{a}{2} + \sqrt{-\frac{a^2}{4} + \frac{1}{27}}} + \sqrt[3]{-\frac{a}{2} - \sqrt{-\frac{a^2}{4} + \frac{1}{27}}}



Also shoutout to the admin. Some fan of Lukyon on soyjak.sharty thinks that you are the good person to discover if manga Lukyon is a virgin. A bit silly question, but he really begged me to ask it. :(
>>
  No. 633 >>638
>>631
I've never thought math in the toilet, but this morning I tried it and I found a root for radicals of (2k+1)th-degree equations, THANKS dude! I hope in tomorrow. I'm not OP.
>>
  No. 638
>>633
To be honest talking about pooping is a bit cringe for me, but you won't get bored on a toilet if you do maths. :)
>>
  No. 639 >>641
Also I had error in >>631

I left minus after squaring and the end should be like this:

"Then we solve
qq
:

1+3a24−q23=01 + 3\sqrt[3]{\frac{a^2}{4} - q^2} = 0


a24−q23=−13\sqrt[3]{\frac{a^2}{4} - q^2} = -\frac{1}{3}


a24−q2=−127\frac{a^2}{4} - q^2 = -\frac{1}{27}


q2=a24+127q^2 = \frac{a^2}{4} + \frac{1}{27}


q=a24+127q = \sqrt{\frac{a^2}{4} + \frac{1}{27}}


and thus:

x=−a2+a24+1273+−a2−a24+1273 x = \sqrt[3]{-\frac{a}{2} + \sqrt{\frac{a^2}{4} + \frac{1}{27}}} + \sqrt[3]{-\frac{a}{2} - \sqrt{\frac{a^2}{4} + \frac{1}{27}}}



Also shoutout to the admin. Some fan of Lukyon on soyjak.sharty thinks that you are the good person to discover if manga Lukyon is a virgin. A bit silly question, but he really begged me to ask it. :("
>>
  No. 641 >>648
>>639
Why are you complitated your life?
There are at least two different easy ways to solve
x3+px+q=0x^3+px+q=0

One of these is
x=y+ayx=y+\frac{a}{y}

y3+a3y3+(3a+p)(y+ay) y^3+\frac{a^3}{y^3}+(3a+p)\left(y+\frac{a}{y}\right)
+q=0
{3a+p=0⇒a=−p3y6+qy3−p327=0\begin{cases} 3a+p=0\Rightarrow a=-\frac{p}{3} \\y^6+qy^3-\frac{p^3}{27}=0 \end{cases}

y3=−q2∓q24+p327y^3=-\frac{q}{2}\mp \sqrt{\frac{q^2}{4}+ \frac{p^3}{27}}

x=−q2+q24+p3273+−p3−q2+q24+p3273\boxed{x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+ \frac{p^3}{27}}}+\frac{-\frac{p}{3}}{\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+ \frac{p^3}{27}}}}}
>>
  No. 648 >>649
>>641
Because I love reinventing the wheel
>>
  No. 649 >>650
>>648
Emoji Image you are started from solution when you wrote
x=sqrt[3]p+q+sqrt[3]p−q x=sqrt[3]{p+q}+sqrt[3]{p-q}

If you reinvente the wheel how do you solve
x5+px2+qx+t=0 x^5+px^2+qx+t=0
or
x6+ax3+bx2+cx+d=0 x^6+ax^3+bx^2+cx+d=0
? Emoji Image
>>
  No. 650 >>651
>>649
Me again, edit: when you wrote
x=p+q3+p−q3 x=\sqrt[3]{p+q}+\sqrt[3]{p-q}
sorry for the mistakes Emoji Image
>>
  No. 651
>>650
I said that I guessed it. I kinda remembered the outlook of the original solution with the sum of two cubic roots, but I didn't remember what was inside.
>>
  No. 702
AAAAAAAAA

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