A Cubic Functional Equation With Nice Results 😄

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[@coreyyanofsky]

at this point i've done so many of these x ± (1/x) problems that i can damn near do this one by inspection [x - (1/x)]³ = x³ - (1/x)³ - 3[x - (1/x)] x³ - (1/x)³ = [x - (1/x)]³ + 3[x - (1/x)] f(x) = x³ + 3x

[@freddyalvaradamaranon304]

Muchas gracias profesor por compartir un nuevo vídeo de funciones 😊😊. Con cambio de variable y propiedades de ecuaciones cuadraticas. Mi hija y mi persona estamos muy agradecidos por tan buena explicación. ❤😊❤😊.

[@Mathsfighters0707]

Nice 🎉

[@anestismoutafidis4575]

The chain breaks-function suits as follows: f(x-1/x)=x^3-(1/x^3) x^3-1/x^3-1/x^3-1/x^3-1/x^3...±0, if Σ (n•k) with k=(-1/x^3)^-1 goes n•(k1÷k2÷k3....)=0, then (x-1/x) has the limit- value lim x-> ±0 Result: The function of the chain- breaks with f(x-1/x)= lim x->±0 is a converse function, with f(x)=x^3

[@yakupbuyankara5903]

F(x)=x^3+3x

[@jpolowin0]

Start by assuming that the function is cubic, since the "input" is monic and the "output" has a cubic term. Further, the coefficient of the cubic term is 1. So: f(y) = y³ + ay² + by + c. Plug in (x + 1/x), expand, and solve for a (0), b (3), and c (0).

[@scottleung9587]

I used a third method, in which I found expressions for x^2-(1/x^2) and x^3-(1/x^3) in terms of a dummy variable. From there, it was pretty simple - I didn't need any radicals or identities.

[@WahranRai]

2:55 Something is wrong with your change of variable !!! You set t = x- 1/x and then you substitute each x by its value as a function of t inside f(...). NO !!! According to your hypothesis : f(x-1/x) = f(t) !!!

[@Mediterranean81]

f(x-1/x)=x^3-1/x^3 f(x-x^-1)=(x-x^-1)(x^2+x^-2+1) let t = x-x^-1 f(t)=t(x^2+x^-2-2+3) f(t)=t(x-x^-1)^2 + 3t f(t)=t × t^2 + 3t f(t) = t(t^2+3)

[@phill3986]

👍😎👏✌️👏😎👍