all solutions to x^3=1

We will find all the cube roots of 1, namely all the cube roots of unity. The first method is to solve the cubic equation x^3=1 by factoring and the quadratic formula. The second method is to use imaginary numbers and rotation.

0:00 the algebra way
3:31 the complex way

[mrstutoring]

beautiful. just beautiful. won't help me get back with my ex, but that's a different story for a different day.

[adamhrankowski]

I prefer the polar form, because it's so easy to extend it to higher roots. Plus the regular polygons centred on the origin provide such an effective geometric illustration of the fundamental theorem of algebra.

[Baltie]

I am amazed by your fluency in utilizing 2 markers at the same time!

[matthewzuelke]

I prefer the first solution when solving generically but the 2nd solution is a fun alternative that i wouldn't have thought of

[donmoore]

I always liked it when professors showed multiple solutions. Well done.

[juhaszelodmusic]

I graduated two years ago with a BSc in Electrical Engineering and now I simply enjoy your math videos more than ever before when I had to do maths at university 😂 Thank you ❤

[Flamingpaper]

These are all the fourth roots of 1:
1
-1
i
-i

[pbondin]

Thank you for your clear derivation of the cube roots of unity. I also remember my high school teacher (more than 50 years ago!) demonstrating that the cube roots of unity are: 1, w and w^2. From your derivation, it was easy to see that, if you let w = (-1/2 + i*sqrt(3)/2), then w^2 = (1/2 + i*sqrt(3)/2)*(1/2 + i*sqrt(3)/2) = (-1/2 - i*sqrt(3)/2). I know this is obvious to most of your subscribers, but for me it’s another reminder of how much I enjoyed (and still do) mathematics.

[JeanDAVID]

When you multiply by i, you rotate by 90°. Here we have three roots so three rotations with different angles. In fact, for x=1, rotation is 0°, for -1/2+i sqrt3/2, rotation is 120°, for -1/2-i sqrt3/2, rotation is -120° or 240°. Three following rotations from position (1, 0i) with these angles will return to position (1, 0i)

[saxbend]

Bring in the quaternions too, why not? :)

[jeromesnail]

Another method is just to think about 1 as e^(2kπi) for all integer k.
Then ³√1 = (e^(2kπi))^(1/3) = e^(2kπi/3)
If k=0 you get e^(0) = 1
If k=1 you get e^(2πi/3) = cos(2π/3) + isin(2π/3) = -1/2 + i√3/2
If k=2 you get e^(4πi/3) = cos(4π/3) + isin(4π/3) = -1/2 - i√3/2
For k<0 or k>2 you get the same results knowing that for any integer n and any real number x, , cos(x+2nπ) = cos(x) and sin(x+2nπ) = sin(x)

[Fematika]

Fun fact - An Eisenstein integer is a + b * w for integers a and b and w = e^(2pi/3), a cube root of unity. The units in this ring (the numbers that have a multiplicative inverse) are simply just plus or minus the cube roots of unity. Then, you can extend this to what is known as an Eisenstein prime, which is an Eisenstein integer who's only divisors are the units, the cube roots of unity, and themselves.

[Ali_Pxll]

I was studying calculus, & I studied a lot of ur content, then I faced x^3 - 1.. searched on YT and found u.
you're a real math finisher

[jackeea_]

2:26

This is 1

Minus 4 that's -3

Quik mafs

[aahpandasrun]

Thanks for explaining so many details that other videos tend to just skip over, such as "roots of unity means roots of 1"

[PM]

Really liked how you handled the two colors for the markers without making a mistake even once!

[holyshit]

They will appear in formula for roots of cubic equation
I think that the Moivre will be useful later in so called casus irreducibilis
Way for solving cubics can be generalized to quartics

For cubic equation we can use following thing

(A+B)^3=A^3+3A^2B+3AB^2+B^3
(A+B)^3=A^3+B^3+3AB(A+B)
(A+B)^3=3AB(A+B)+(A^3+B^3)
which is in the same form as the equation
y^3=3py+2q, (after shifting variable to eliminate x^2 term)

For quartics we can use the same concept


which is in the same form as the equation
y^4=2py^2+8qy+r, (here we also shift variable to eliminate x^3 term)

I got this identities after playing with symmetric functions

After comparing coefficients we will get system of equations which can be transformed to
Vieta formulas for sextic polynomial reducible to lower degree polynomial
(for cubics sextic resolvent is reducible to quadratic and for quartic sextic resolvent isreducible to cubic)

Vieta's formulas involving elementary symmetric functions of the roots and coefficients of polynomial
can also be useful for solving cubics and quartics

Symmetric function does not change after permutation of its variables

Another way for solving cubics and quartics is

x^3+px+q=A(x+n)^3-B(x+m)^3

[GeekTommy]

Your videos about series and integrals helped me a lot to pass my calculus exam, thank you so much! I'm Italian and we call that "Analisi Matematica 1", it's the exam about limits, derivatives, series and integrals. Even if I'm not English, your explanations are really clear and understandable, I wish all the teachers were like you! Keep up the great work! A question, will you ever make videos about linear algebra (matrices and so on) and probability stuff?

[jacoblund]

Hey, great video! When I was first learning about the roots of unity there was an exercise which asked me to prove that they sum to 0, and Id like to share my answer with you:
Where w is an nth root of unity
and I'm going to call the series:
1 + w + w^2 +... w^n-1= S
because w^n = 1:
1 + w + w^2 +... w^n-1= w^n+ w + w^2 +... w^n-1
therefore: S = wS
If assume S is not 0 and dicode both sides by S, we get w = 1 which is a contradiction for all n other than 1 therefore S = 0

Thanks for reading!

[aeuhgv]

you could also do:

x^3=1
x^3=e^(k * 2pi * i) with k an integer
x=e^(k * 2pi * i)^1/3
x=e^(k * 1/3 * 2pi * i)
x=e^(k * 2/3pi * i)

k=0, k=1 and k=2 give respectively

x=e^0 x=e^(2/3pi * i) x=e^(4/3pi * i)

you know that 1/3pi rad = 30°
so 2/3pi rad and 4/3pi rad are just mutiples of 30°
Every found number on the unit circle forms a special right angle triangle with one 30° angle
With the unit circle and special right angle triangles you find that

x=1 x= -1/2 + 1/2 * sqrt(3)i x= -1/2 - 1/2 *sqrt(3)i