anyone smart enought?

[woeful jackal]

Solve the equation

[mild meadow]

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[trail gust]

im not smart enough but i can solve it 💔

using SUBSTITUTION works.... find $x=y^2(y-1)$

[abstract bluff]

wolfqz

[woeful jackal]

I am been trying to solve it for 5h

@trail gust i mean and whats next?

Tryed this too

[trail gust]

oh

hi

i forgot

[trail gust]

well put this instead of x in the second equation

wait thats long and tidious work

[woeful jackal]

Yep

[trail gust]

$y=x^2(x-1)=(y^2(y-1))(y^2(y-1)-1)$

[abstract bluff]

wolfqz

[trail gust]

$y=(y^2(y-1))(y^2(y-1)-1)$

[abstract bluff]

wolfqz

[trail gust]

this is some work, but it is solvable

[woeful jackal]

Wait where did you put

The y

In which 1st or second

[trail gust]

we had $x=y^2(y-1)$

[abstract bluff]

wolfqz

[trail gust]

we put it in the second equation after a bit of rearranging

[woeful jackal]

After a bit i got

[young blaze]

If you take a look at both equations, they are the same equations just with x and y reversed

Since $x = y^2*(y-1)$
then also $y = x^2*(x-1)$

[abstract bluff]

muerte

[young blaze]

he simply inserted $x = y^2(y-1)$ into $y = x^2(x-1)$

[abstract bluff]

muerte

[young blaze]

thats how he got this

[woeful jackal]

Oh ok ill go try

Whats next?

[young blaze]

@woeful jackal

hello

[woeful jackal]

Yes

[young blaze]

can u translate

pls

[woeful jackal]

Solve the equation

The translation

Thats all

😀

[young blaze]

oh

[woeful jackal]

Yep

[young blaze]

what have u done

so far

[woeful jackal]

This is the last

Tryed everything

Thats what i did

[young blaze]

$x = y(y^{2}-y)\
y = x(x^{2} -x)$

[woeful jackal]

I did that

Already

[abstract bluff]

Gary

[woeful jackal]

Did that too

[young blaze]

$x = x(x^{2} -x)(( x(x^{2} -x))^{2}- x(x^{2} -x))$

[abstract bluff]

Gary

[young blaze]

if i did things correctly

this will give the relation

xy = 1

NAH

wait

[woeful jackal]

Not an easy one

I would say

😃

A”

[young blaze]

Didnt see it earlier

Is it solved?

[woeful jackal]

Nope

[young blaze]

Oh boy let me see 😄

[woeful jackal]

I am so lost in x and y

5 pages of my book is full of x’s and y’s

[vernal bough]

Uhh u don't have to over complicate stuff , see both the equations they are both symmetrical , so in this case u say x= y for the solution
So put x in place of y. And solve the cubic equation u get

If u are having problem solving cubic equation then ping

[woeful jackal]

I did that for both equations and got

[woeful jackal]

.

[woeful jackal]

.

[vernal bough]

No not that

Am telling to simply put y = x in eq (1)

And solve the cubic since both equation are symmetrical

Uhh I will just show in paper

[young blaze]

I haven't considered this, but right there should be no possible solutions with x unequal y

that makes it a lot easier

ah

x = -y

its comin

if i did it correctly

lmao

Would suprise me

[vernal bough]

Nah

It's just , only and only x= y

[young blaze]

if x = -y, then $-y^3 = y^2 + y$ ?

[abstract bluff]

muerte

[young blaze]

yeah that seems a bit weird ngl

[vernal bough]

Mhm

[young blaze]

this is it

I was too worried that there might be a solution with y unequal x

hm yes

x = y

is the one

i put negative by mistake

doin the calculation

lmao

[vernal bough]

Hm true, I guess u can confirm it by putting this graphs in desmos

[young blaze]

$y^{2}(x^{2}-x) - y(x^{2}-x) - 1 = 0$

[abstract bluff]

Gary

[vernal bough]

Even tho I bet there's no value for x unequal y

[young blaze]

surely this will lead somewhere

mine is like

[vernal bough]

hell nah

[young blaze]

3 page

lmao

[vernal bough]

Mine in 1/2 page

[young blaze]

just solve the equation for x, mathjack 💀

$x^{2}(y^{2}-x) - x(y^{2}-x) - 1 = 0\
y^{2}(x^{2}-x) - y(x^{2}-x) - 1 = 0$

[abstract bluff]

Gary

[young blaze]

i solved these two

to get the

relation x = y

[vernal bough]

Dayum bro

Respect for proving it

[woeful jackal]

Ok ill need to read

Out everything

Now

😭

[young blaze]

just read what @vernal bough said

[young blaze]

he solved it right here

[woeful jackal]

@vernal bough thanks

[viscid folio]

@woeful jackal has given 1 rep to @vernal bough

[vernal bough]

Np brother , u can close this channel if no other doubts

[woeful jackal]

I went to the store ill recheck it again and then close it

[vernal bough]

Sure

[young blaze]

Based af

x = y

Lmao

[woeful jackal]

XD

[vernal bough]

💀nah I feel ur pain brow , sometimes I fill up millions pages just to realize it was of just one line

[woeful jackal]

Yea

I did that with my math

Book

[iron hull]

what about

they meet along the axis of symmetry.

so x=y be the only sol.

[young blaze]

Wait how u see that

[iron hull]

because that's how these curves work

i.e. cubics

[young blaze]

Oh yeh

So there

Graphs are similar

[iron hull]

if we have f(n) = n^3 - n^2
then the 2 curves we have are :
y = f(x)
x = f(y)

[young blaze]

This does show similarities

Kinda

[iron hull]

and as it's a cubic they just meet at the axis of symmetry i.e x=y

just a polynomial thing

[vernal bough]

Hmm yea only in the case of symmetric curves whose we wanna find f(x) = f^-1(x)

[iron hull]

mhm

[vernal bough]

Since f^(-1)(x) is always mirror image of f(x) along y= x , they only meet at y = x