Algebra is the king of mathematics. I wish I truly spent time developing that aspect of my math before calculus and other things showed up.

Guys look at my cool millionth degree polynomial: x¹⁰⁰⁰⁰⁰⁰ = x¹⁰⁰⁰⁰⁰⁰ + x-1 😂

10:39
"Those who stop learning, stop living"
Is that a threat?

Technically it’s a hexic (or sextic??) equation as the x^7 on both sides cancel, which means there should be 6 roots in C including multiplicity, as u found

The only septic I can solve is figuring out what happens when I flush my toilet lol.

I suppose sanitary engineers need to solve septic equations...

Pascal Triangle 📐

The easy way to memorize 49 times 7 is 50 times 7 is 350 and minus 7 is 343

I love your English pronunciation. I can watch your videos with 1.5 speed and completely understand your lecture.
Hello from Russia!

I watched this video twice because I like watching you solve problems like this.

Nice problem. Note that all of your solutions are multiples of 7: 0*7,-1*7,w*7 and (w^2)*7 where  w and w^2 are complex cube roots of unity. This corresponds to your factorization.

The complex solutions can be presented as (7/2)*e^(i*2*pi/3) and (7/2)*e^(i*4*pi/3).

Very interesting video

I would have to go check my old abstract algebra textbooks to find the exact way it's described, but if I remember correctly, for any prime p, you have (x+y)^p=x^p+y^p for all x and y when considering over the field Z_p.  The trick is to realize that for a prime, all the binomial coefficients in the expansion of the left hand side are a multiple of p, except the first and last (which are always one).  Since p~0 in that field, all the extra terms simply disappear.  Granted, the solution over the complex numbers given here is the best interpretation of the problem when given without a more specific context, but it's nice to know that there really is a context where the naive student's thought that (x+y)^2=x^2 + y^2 actually does hold.

Cool way to expand! You are a very good teacher.

U can say complex solutions....
Anyway very informative 😁😁

I understand you did solve this equation using elementary and tricky (!) calculations.
Some of them I had never heard of…! However, I found an other way and I wanted to share it with you.
I suggest the following : let us replace x / 7 by y, then comes  (y + 1)^7 = y^7 + 1. (Then we get rid of the 7^7 which slows down the path…).
Let us talk about P(y) = (y + 1)^7 - y^7 - 1. Obviously, P(0) = 0, then y divides P.  Again, P(-1) = 0, then (y+1) divides P.
As we notice that d°P = 6, it is a good idea to test if a third or sixth root of unity is a solution of P (y) = 0. Let’s try j = exp (2 i π/3) :
P (j) = (j+1)^7 + j^7 + 1 = - j^14 - j - 1 = 0 (because j^3 = 1 and 1 + j + j^2 = 0).
Then y - j divides P, and also y - j^2, and also (y - j) (y - j^2) = y^2 + y + 1.
We could now divide P by y (y+1) (y^2 + y + 1)… but I think this is better :
P’ (y) = 7 (y+1)^6 - 7 y^6, so that P’ ( j ) = 7 (1 + j)^6 - 7 j^6 = 7 (j^12) - 7 = 0.
Then (y - j)^2 divides P, and also (y - j^2)^2, and also  (y^2 + y + 1)^2.
As the highest term of P is 7 y^6, we conclude that P (y) = 7 y (y+1) (y^2 + y + 1)^2,
And the solutions are 0, -1, j and j^2, these last two being double.
Multiplying by 7 give back x = 0, -7, 7j and 7 j^2, these last two being double.
Thanks for your very interesting videos. (I love your channel !). 😊

Awesome. Very good Explanation.

Can anyone please explain why the imaginary solutions are written twice?

Septic equation... sounds like something that needs to be pumped out... 😅
But seriously... this is really great stuff!
I figured out quickly that x = 0 and x = -7 were roots... then got stuck on the rest. Thanks for showing me the trick!