it's more simple if you observe that a-b=1 then you know (a-b)^2=1 and you have a^3-b^3=1 . Finally you find ab=0. The next is simple

😊😊

Muy bien.

You told that solition very well.Thank you.

It just need patience

Eksi....😂

let y=(x-3)/(x^2+3) and note that (x^2+x)/(x^2+3) = 1+y (use synthetic division if this is not obvious). Problem maybe written (1+y)^3-y^3-1=0 expand to obtain 3*y^2+3*y= 0 , y=0 or y=-1; y=0=> x=3, & y=-1 => x^2+x=0 or x = 0, or x=-1,

Если присмотреться то если из первого числителя отнять второй равно знаменателю, это не зря

How about the same equation but make it functional?

Un truc que je voudrais voir sur tes vidéos, c’est le domaine de définition. Ici par exemple : X^2-3≠0 ===> Df = R - { -√3 ; √3}

please more equation like this ! I love this so much

Nice problem

Another method: The original equation is: ((x^2 + x) / (x^2 + 3))^3 - ((x - 3) / (x^2 + 3))^3 = 1 (1) This equation may be written as: a^3 + b^3 + c^3 = 0 , denoting a, b, c as: a = (x^2 + x) / (x^2 + 3) , b = - (x - 3) / (x^2 + 3) , c = -1 Is easy to show that: a + b + c = 0 and in this case: a^3 + b^3 + c^3 = 3abc = 0 . Therefore, eq. (1) reduces to: ((x^2 + x) / (x^2 + 3)) ((x - 3) / (x^2 + 3)) = 0 (2) However, x^2 + 3 ≠ 0 and therefore eq. (2) reduces to: x (x + 1) (x - 3) = 0 which gives 3 roots: x=0 , x = - 1 , x = 3

asnwer=x1 isit

Kwerage tight ywello wy werji x aydelem xsquaru yishalegnal middre chikko azawoch ekhkhkhkheeey yihe dirty ymiyasetella bahriyachun bmin y9gelets y9hon ainstien rasu binor qala5 yattalet nbere kirfatamoch