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.

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

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

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

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

Pascal Triangle 📐

Since the imaginary solutions get squared, you should also be able to use the negative of those imaginary solutions. Thus, the four imaginary solutions should be [+/-7 +/- i sqrt(3)]/2. Note the plus or minus in front of the 7. The other two (real) solutions are x = 0 and x = -7, as you noted.

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

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.

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.

Nice math solution.. I see you video everyday. It is really so helpful for me. Thank you my Boss. Mahin From Bangladesh.

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

This is definetly an algebra's student dream.

I think i have a general solution to this kind of equation: (x+n)^n = x^n + n^n ( n is natural number, n > 1). Divide both side of equation by n^n. We will have (x+n)^n / n^n = x^n / n^n + 1 which is equivalent to (x/n + 1)^n = (x/n)^n + 1. Let t = x/n, then the equation will become (t+1)^n = t^n + 1. So now we will focus on solving t It is easy to see that if n is even then we just have one solution is t = 0 and if n is odd then t = -1 or t = 0. The main idea here is show that these are only solutions. So let f(t) = (t+1)^n - t^n - 1 Case 1: n is even f'(t) = n.(t+1)^(n-1) - n.t^(n-1) f'(t) = 0 (t+1)^(n-1) = t^(n-1) Notice that n is even so n-1 is odd. Then we have t+1 = t (nonsense) So f'(t) > 0. Thus f(t) = 0 has maximum one solution. And t = 0 is the only solution here. Case 2: n is odd. We have f''(t) = n(n-1).(t+1)^(n-2) - n(n-1).t^(n-2) f"(t) = 0 (t+1)^(n-2) = t^(n-2) Notice that n is odd so n-2 is odd Then we have t+1 = t (nonsense again) So f"(t) > 0 which leads us to the fact that f(t) = 0 has maximum two solutions. And t = 0 and t = -1 are two solutions. After we have solved for t, we can easily solve for x.

Bro u got to be the best Maths teacher

That s a rly cool explanation but the third is wrong to me : if ( x2 + 7x + 49 )2 equals 0 then x2 + 7x + 49 equals square root of 0 so 0 and x2 + 7x + 49 is ( x + 7 )2 so we replace and then we take out the square of ( x + 7 )2 so x + 7 = 0 and we get the same answer than the last one

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

What is the simplification of (x+y)^n -x^n -y^n??

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

Why couldn't we use the Pascal triangle for the first part (x+7)^7 ?

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

is there a general formula for factoring (x+y)^(2n-1) - x^(2n-1) - y^(2n-1)

Because x^2+7x+49 is squared can -x^2-7x-49 be used to solve for two other roots rather than repeat?

(x+y)^7-x^7-y^7=7xy(x+y)(x^2+xy+y^2)^2　　; why (x^2+xy+y^2)^2　It's not a math formula, but there's no explanation.

Eulers equation (a +b)^n= a^n+ b^n....for n=1,2....

Its more like a hexic (is that the word for 6?) Rather than septic because the x⁷ terms cancel each other

Interesting to do the factoring, I'll try. But I'm curious how such things are obtained, I'd guess that can be done by synthetic division , if you have a clue what to obtain at tĥe end. Not quite obvious. Especially with the 7th degree, that incomplete square squared, looks overwhelming, I'd say 😅

I generalised this for n is odd. Tried doing it for n is even and couldn't get anywhere 😩 Solve for x in terms of n if (x + n)^n = x^n + n^n and n ∈ Z^+. Case 1 - n = 1 : →x + n = x + n There are no valid solutions for x. Case 2 - n is odd and n ≥ 3: →(x + n)^n - x^n - n^n = 0 After looking at n = 3, 5, 7 and so on, we notice a pattern: →(n^2)*x*(x + n)*(x^2 + nx + n^2 )^((n - 3)/2) = 0 →x = 0, x = -n For x^2 + nx + n^2 = 0 , where n > 3: →x = (-n ± √(n^2 - 4n^2 ))/2 →x = (-n ± n√3*i)/2 If anyone can provide a generalisation for n is even, then please reply to my comment 😊

I really like this one!

I don't like how you have to explain how to solve 49x=0 but not how to simplify (x+7)^7-x^7-7^7. Here's how I would have done it. Being able to cancel out the x^7 and constant terms is too good, so I would expand the polynomial. Don't want the coefficients blowing up? Substitute x=7t and the problem reduces to (t+1)^7=t^7-1. After the expansion, subtraction, and division by 7, we get t^6+3t^5+5t^4+5t^3+3t^2+t. Factor out the t, and factor the rest by grouping terms with the same coefficients. t^5+1+3t(t^3+1)+5t^2(t+1)=(t+1)(t^4-t^3+t^2-t+1+3t^3-3t^2+3t+5t^2)=(t+1)(t^4+2t^3+3t^2+2t+1)=(t+1)(t^2+t+1)^2. Solve for t, multiply by 7 to get x.

Its true for all x in Z/7Z

I actually got the first and last term thing right, I just didnt know how to get the numbers in the middle lol

Amazing

Would your experience solving this septic equation qualify you to repair our nasty, leaky, smelly septic tank? Nice job on choosing a relatively obscure term like septic as it could possibly enhance Search Engine Optimization(SEO), resulting in more page views from wordsmiths!

Septic ?

I wouldve just said by fermas last theorem x can only be equal to 0

isn't that equation more simple using pascal triangle ?

Tús es o cara. Thank you

Fermat conjectures

At the very end you say that 49 - 4(49) is negative 3 but it's negative 3(49) aka 147

A septic equation turned into a sextic equation..... I never thought that algebra so "dirty".

It is easy to guess the two solutions x= 0, x = -7 , but one has to show that these are the only real solutions.

The 7th root is ∞.

x = 0 ez

put some TCP on it !!!😅😃

I have a septic infection 😂

why only six answers? shouldn't there be seven?

Believe it or not, I have made a summation for this exact problem but for all n not just 7

Couldnt you just 7th root the entire equation and have all the exponents cancel out?

(x+7)^7=x^7+7^7 (0+7)^7=0^7+7^7 7^7=7^77=7 x=0

septic… hm… 😂

To me its clear at the start that x must be less than 1.

X=0

multiplicity solution at end has been repeated

x=0

0

Don’t do it like this just brute force it and then synthetic Devine it Trust me man trust me

Medical person me reads “septic” 🤒

( x + 7 )^7 - ( x^7 + 7^7 ) = 0 49 x ( x + 7 )( x^2 + 7 x + 49 )^2 = 0 x = { 0 , - 7 , 7 w , 7 w^2 } x^3 - 1 = ( x - 1 )( x^2 + x + 1 ) = 0 x = { 1 , w , w^2 } , w € C , w^3 = 1 😊🤪👍👋

Mei muslman hon hindu nhi hon

Do me a favour: Don't call the non-real solutions 'imaginary'! They are called 'complex', ok. Nevertheless, the 'number' i ist called the 'imaginary entity'. Furthermore, there are 'imaginary numbers'. These are complex numbers without a real part or having zero as real part respectivly.