What do you think about this problem? If you're reading this ❤️. Hello My Friend ! Welcome to my channel. I really appreciate it! @higher_mathematics #maths #math
If you naively take the cube root of the original equation & arrive at x=x-1, there's a nominal solution. Namely, |x|=∞, b/c then x-1 is the same as x (you can't subtract infinity from both sides to get 0=-1). Of course, this depends on whether you wanna allow for infinity. To be more precise, for complex x, you can take the cube root of both sides of x³=(x-1)³ if you take into account the argument of x. Namely, x·e^(2πin/3)=x-1 for n=0,±1 ⇒ x=1/[1-e^(2πin/3)]. For n=±1 you get the 2 complex roots in the video. But for n=0, x=lim_{ε⇾0} 1/[1-cos(ε)-isin(ε)]=lim_{ε⇾0} 1/(-iε)=±i∞.
This video is one of the many videos about this theme. Why not solve it generally? x^n = (x+a)^n Solution: x_k = a/(e(n)^k-1), where e(n) is the nth primitive root of unity and k=1, 2, ..., n-1
Perhaps a good topic would be the domain of various identities and rules. E.g. when is sqrta•sqrtb = sqrt(ab)? If a and b are both negative this will go wrong.
If you naively take the cube root of the original equation & arrive at x=x-1, there's a nominal solution. Namely, |x|=∞, b/c then x-1 is the same as x (you can't subtract infinity from both sides to get 0=-1). Of course, this depends on whether you wanna allow for infinity.
To be more precise, for complex x, you can take the cube root of both sides of
x³=(x-1)³ if you take into account the argument of x. Namely,
x·e^(2πin/3)=x-1 for n=0,±1 ⇒ x=1/[1-e^(2πin/3)].
For n=±1 you get the 2 complex roots in the video. But for n=0,
x=lim_{ε⇾0} 1/[1-cos(ε)-isin(ε)]=lim_{ε⇾0} 1/(-iε)=±i∞.
Actually... We always do "Rationalize the denominator" . But you did the opposite.
Absolutely right
This video is one of the many videos about this theme. Why not solve it generally?
x^n = (x+a)^n
Solution: x_k = a/(e(n)^k-1), where e(n) is the nth primitive root of unity and k=1, 2, ..., n-1
x³ = (x - 1)³
x³ = (x - 1)².(x - 1)
x³ = (x² - 2x + 1).(x - 1)
x³ = x³ - x² - 2x² + 2x + x - 1
0 = - 3x² + 3x - 1
3x² - 3x = - 1
x² - x = - 1/3
x² - x + (1/2)² = - (1/3) + (1/2)²
[x - (1/2)]² = - 1/12
[x - (1/2)]² = - i²/12
x - (1/2) = ± i/√12
x - (1/2) = ± (i√12)/12 → recall: √12 = √(4 * 3) = 2√3
x - (1/2) = ± (2i√3)/12
x - (1/2) = ± (i√3)/6
x = (1/2) ± [(i√3)/6]
x = (3 ± i√3)/6
Perhaps a good topic would be the domain of various identities and rules. E.g. when is sqrta•sqrtb = sqrt(ab)? If a and b are both negative this will go wrong.
An idea.(1-1/x)^3=1 so 1-1/x = 1, w, w^2. (w is omega)
I tried OpenAI and it found also x=1 as a solution.
After few trials it accepted the error
(x-1)^3=x^3-3x^2+3x-1 , x^3=x^3-3x^2+3x-1 , -x3 and *(-1) , 3x^2-3x-1=0 , /(3) , x^2-x+1/3=0 , x=(1 |+/-| sqrt(1-4/3))/2 ,
x= (1+i*sqrt(1/3)/2 , (1-i*sqrt(1/3)/2 ,
GREAT VIDEO! Liked and subscribed ❤
What about x^3=(x+1)^3?
There's would be also two complex roots, but with a negative sign: -1/2 +- i*sqrt(3)/6
Substitute y=-x and it reduces to the original eqn.