Solving a septic equation

[@SpiroGirah]

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.

[@ernestdecsi5913]

I am 70 years old and I am just now realising how much I have always been interested in mathematics. It is a pity that when I was young, KZbin did not exist and the beauty of mathematics was not so visible.

[@MrJasbur1]

Yeah, except when algebra has a rule that says that you have to pretend that an equation has more solutions than it does because of multiplicities. They should get rid of that rule. Imaginary numbers may be useful, but I’m not sold on multiplicities being the same.

[@pedrogarcia8706]

@@MrJasbur1 it's not that deep. multiplicity just means when you factor the polynomial, the factor is written twice. for all intents and purposes, the equation has 4 solutions, but it still has 6 factors, 2 of them just appear twice.

[@SalmonForYourLuck]

​@@pedrogarcia8706So that's why he wrote the Imaginery solutions twice?

[@pedrogarcia8706]

@@SalmonForYourLuck yeah exactly, if you were to write the factorization of the polynomial, the factors would be (x minus each solution) and the solutions with multiplicity would be repeated. You could also write those factors squared to only have to write them once.

[@mac_bomber3521]

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

[@PrimeNewtons]

Only if you feel threatened.

[@mcvoid7052]

Better get to learning.

[@MangoMan1963]

"Those who start learning, stop living" ~Avg JEE/NEET aspirant

[@t-seriesgaming7408]

😢😢​@@MangoMan1963

[@adw1z]

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

[@dayingale3231]

Yesss

[@alwayschill4522]

yeah i saw that too... its giving clickbait just kidding we love!

[@erenshaw]

Thank u I was so confused in why there was only 6 solutions

[@plutothetutor1660]

Factoring an x leads to a quintic equation too!

[@shinjonmal8936]

I write it as "Hectic Roots" because it is indeed hectic to find them

[@autolightview]

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

[@dougaugustine4075]

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

[@wavingbuddy3535]

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

[@Simpson17866]

I just solved it in my head :D

[@adw1z]

@@Simpson17866 sorry to be a killjoy but ur polynomial is technically 1 degree only 😭

[@Simpson17866]

@@adw1z ... That's the joke.

[@the-boy-who-lived]

After hours of work through trials and errors and using qudralliontic equation and almost proving Riemann hypothesis, I figured out it is 1-x=0

[@AverageKopite]

@@the-boy-who-lived👏👏🙌😂

[@mahinnazu5455]

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

[@mahinnazu5455]

Sir I hope u can support me to learn Mathematics.I love to do Maths.

[@Blaqjaqshellaq]

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

[@bobbun9630]

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.

[@echandler]

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.

[@donwald3436]

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

[@PrimeNewtons]

Now you have one more

[@wayneosaur]

I used the binomial expansion of the middle part to get 7^2x(x^5 + 21x^4 + 35*7*x^3 + 35*7^2*x^2 + 21*7^3*x + 7^5) = 0. By symmetry arguments you can see that this is solved when x^3 = -7^3 which is equivalent to (x/-7)^3 = 1. The three roots of unity are {1, -1/2 + sqrt(3)/2 * i, -1/2 - sqrt(3)/2 * i}. Set x/-7 to these three values you get the last 3 roots.

[@dmk_5736]

if you set to middle point y - 3.5 = x , you would get only even powers of y, so z=y², and you would get third degree equation, after division by (z - 3.5²) you would get perfect square (z+3*3.5²)², i.e. (z-3.5²)*(z+3*3.5²)²=0 is the same equation taking into account z=y², y=x+3.5.

[@kornelviktor6985]

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

[@NotNochos]

Or do 28^2 - 21^2

[@15121960100]

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

[@kdipakj]

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

[@souverain1er]

Prof, Wouldn’t it be easier to just use pascal triangle for the expansion?

[@VARUNKHANANI]

yes, even the binomial theorem can do the work

[@VARUNKHANANI]

binomial theorem is the SUITABLE one cuz remembering pascal's triangle can be a hectic task(my opinion)

[@trankiennang]

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.

[@knownuser_bs]

also good way to solve brother

[@himadrikhanra7463]

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

[@5Stars49]

Pascal Triangle 📐

[@Siraj-123-q5p]

That's a lengthy process because power is too big (7)

[@nmanoharreddy2525]

Can we use binomial expansion....

[@AbulfazMehdizada-q4t]

(x+a)^n = x^n + a^n, if n=2k, x=0. if n=2k+1, x=0 and x=-a

[@sajuvasu]

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

[@roufaidbelmokh6087]

we could do it another way? like we know that the sloutions are imaginary so we suppose that x=a+bi and use euler formula and try to find a and b ? i didnt try it yet i am just lerning so my way is doable or no?

[@pojuantsalo3475]

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

[@PrimeNewtons]

😂

[@遠傳五華]

X=-7 is an obvious root, which make X+7=0. Any other root, X+70, then we could divide both side by X+7 and apply with substitution, leads to 1=A^7+B°7 and 1=A+B. Using Pascal's Triangle and some calculations, we get A=0,(1+-i√3)/2, then get 3 roots of X.

[@lukaskamin755]

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 😅

[@Christ-The-Lord-92]

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

[@joshuaharper372]

But (x+7)²=x²+14x+49

[@ElAleXeX]

Could this mean we can express (x+y)^n as x^n + nxy(x+y)(x²+xy+y²)^((x-3)/2) + y^n where n is an odd positive integer?

[@tommc1425]

The pattern in the video falls apart after n=7 I'm afraid. You can substitute in x=y=1 to see that it doesn't equate at high values of n

[@Viaz1]

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

[@tebourbi]

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

[@ThenSaidHeUntoThem]

Sextic

[@AlexMarkin-w6c]

Alternative Solution for real roots only. Consider the function f(x)=(x+7)^7 - (x^7+7^7) First, compute the derivative: f'(x)=7(x+7)^6-7x^6 Setting the derivative to zero to find critical points: f'(x)=0 (x+7)^6=7x^6 Taking the sixth root on both sides: |x+7|=|x|. This implies x=-3.5, which is the only extremum and minimum point of the function. Since f(x) monotonic and continuous, it intersects the x-axis twice. Additionally, x=-3.5 is the axis of symmetry for the function derived from the binomial expansion. Given this symmetry, the second solution is 3.5 units away in the negative direction from the axis of symmetry at x=-3.5, which gives x=-7. Therefore, the real solutions are x=-7, x=0.

[@jjjilani9634]

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

[@thecrazzxz3383]

You mean Newton's binomial probably

[@thecrazzxz3383]

You don't necesseraly use Pascal's triangle to develop binomials, there's a formula for the binomial coefficient

[@dmk_5736]

equivalent to ((x+3.5)²-3.5²)*((x+3.5)²+3*3.5²)²=0 (got here by replacing variables.)

[@AbulfazMehdizada-q4t]

we can solve this problem by sketching the graph, where we will see that that they intersect in the point of 0

[@ernestdecsi5913]

I really like this one!

[@RyanLewis-Johnson-wq6xs]

(X+7)^7=X^7+7^7 X=-7 ,X=0,X=(-7±7Sqrt[3]i)/2=-3.5±3.5Sqrt[3]i

[@zyklos229]

I would say (x + 7)^7 = sum i=0..7 binomial (7 over i) x^i 7^(7-i) Leaving us with 1 x=0 solution and polynomial of degree 5 equals 0, so 5 more solutions, none of it positive. (-7) seems a solution, dividing leaves us with solveable 4th degree. The shown factorization makes sense, but appears little bit abitrary 🤔

[@maburwanemokoena7117]

This is definetly an algebra's student dream.

[@mathyyys8467]

Its true for all x in Z/7Z

[@dujas2]

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.

[@荻野憲一-p7o]

一見 7 次方程式に見えるけど、 x = 0 を分離すれば 5 次で しかも相反方程式だから実質 3 次。 それが、 ほぼ自明な解 x = -7 を使って因数分解すれば、 1 次と 2 次に分かれる。 日本の高校では、教科書の例題レベルの問題。

[@akshatbhatnagar9333]

why not just expand using binomial and then cancel out the x⁷ and 7⁷ terms you can factor it out afterwards easily....

[@tardisman602]

Those equations were equivalent to 3xy(x+y)(x^2+xy+y^2)^0 5xy(x+y)(x^2+xy+y^2)^1 7xy(x+y)(x^2+xy+y^2)^2

[@tardisman602]

So you can get this equation from (x+y)^a - x^a - y^a = axy(x+y)(x^2+xy+y^2)^b b = (a-3)/2 For odd numbers of a only

[@tassiedevil2200]

​@tardisman602 Yes for a= 3,5,7 but just a caution that is is not so simple in general i.e. for higher values. RHS of your formula is not a correct factorisation of the LHS for a > 7, even though for all positive odd integers a >5 and not a multiple of 3, the roots -1/2 +/- i sqrt[3]/2 do return, either as simple or as double (repeated) roots - but other much complicated sets of roots also arise, not the high level of degeneracy suggested if your formula held for all odd a.

[@MyOneFiftiethOfADollar]

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!

[@PrimeNewtons]

I use that knowledge to fix my septic tank too 😂

[@hayn10]

Septic ?

[@frozenicetea3494]

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

[@matheusespalaor1757]

Amazing

[@ayaansajjad6855]

isn't that equation more simple using pascal triangle ?

[@FishSticker]

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

[@ThePayner11]

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 😊

[@PaulMutser]

Surely for case 1, all values of x are valid solutions?

[@someperson188]

Your formula: (x+n)^n - x^n - n^n = (n^2)*x*(x + n)*(x^2 + nx + n^2 )^((n - 3)/2) doesn't work when n = 9. It does work for n =3, 5, 7. I used Symbolab to compare ((x+9)^9 - x^9 - 9^9)/(81x(x+9)) and (x^2 + 9x + 81)^3. Symbolab says they are different sextic polynomials. I was too lazy to do the calculation by hand.

[@marcelo372]

Tús es o cara. Thank you

[@thecrazzxz3383]

For x in Z/7Z, we always have the equality : (x+7)^7 = x^7 + 7^7

[@thecrazzxz3383]

For those who are wondering, we can prove that with the little fermat's theorem (that states that for all prime number p, all x € Z, x^p ≡ x mod p) Just for those who don't know, the ring Z/7Z is just the set of remainders from 0 to 6 with the addition, multiplication mod 7 To simplify things, saying "For all x in Z/7Z, we always have the equality : (x+7)^7 = x^7 + 7^7" is exactly equivalent to "For all x € Z, (x+7)^7 ≡ x^7 + 7^7 mod 7" In fact, you can prove more generally with little fermat's theorem this lemma : "For all prime number p, for all a, b € Z, (a+b)^p ≡ a^p + b^p mod n" The demonstration is really simple : Let p be a prime number and a, b € Z By fermat's little theorem : (a+b)^p ≡ a + b mod p ≡ a^p + b^p mod p, still by fermat's little theorem

[@mitadas9961]

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

[@sadeqirfan5582]

But what is the point of repeating it if the two repetitions are the same?

[@timothybohdan7415]

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 the speaker noted.

[@JohnNess7730]

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

[@miya-w2o]

(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.

[@renesperb]

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

[@timothybohdan7415]

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.

[@edouardbinet7893]

Fermat conjectures

[@jceepf]

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

[@Coyto3]

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

[@PrimeNewtons]

I would be glad if you can share 😀

[@antonionavarro1000]

¿Lo has demostrado solo para los n impares? ¿Has demostrado lo siguiente?: Si n es un número natural impar, es decir, n=2m+1, con m un número natural cualquiera, se debe cumplir que (a+b)^{2m+1}- ( a^{2m+1} + b^{2m+1} ) = (2m+1) • (a+b) • (a^2+ab+b^2)^{2m-2} Por favor, escribe la demostración. Sería de agradecer que lo hicieras.

[@Coyto3]

@@PrimeNewtons I would have to send you the picture. I wrote it out on my board. I think it has one slight error that I need to fix. I can probably send it in a desmos link.

[@sea1865]

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

[@harley_2305]

That doesn’t work because on the right hand side you have x^7 + 7^7. You can’t take a root in this form because that would basically be saying root(x+y) = root(x) + root(y) and we can test that doesn’t work by just plugging in numbers such as 4 and 5. root(4 + 5) = 3 but root(4) + root(5) ≈ 4.236 so by counter example the root of the sums is not equal to the sum of the roots hence you can’t cancel out powers of individual terms by taking the root of the whole thing, the whole thing would need to be raised to a power for you to be able to if that makes sense. Sorry if this didn’t explain it well

[@tobybartels8426]

The 7th root is ∞.

[@aashsyed1277]

it actually simplifies to a sixtic

[@zweisteinya]

Septic is right

[@marksandsmith6778]

put some TCP on it !!!😅😃

[@rishavsedhain8547]

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

[@glorfindel75]

the starting equation is sixth degree: it has 6 solutions, not seven

[@orenfivel6247]

*complex sol's

[@ilafya]

Allô thé first équation is à polynôme of 6th degre it should has 6 roots and you given only four maybe there are two roots missing🎉

[@tassiedevil2200]

@ilafya - the complex roots are each double (repeated) roots - due to the quadratic they solve being squared in the factorized expression.

[@noblearmy567]

I have a septic infection 😂

[@PrimeNewtons]

😂

[@anestismoutafidis4575]

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

[@JSSTyger]

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

[@JSSTyger]

The reason I say this is that (x+7)^7 = x^7+7^7+positive number, which is greater than x^7+7^7. So really, I could also argue that x can't even be greater than 0.

[@williamdragon1023]

x = 0 ez

[@Bertin-q3y]

X=0

[@dankestlynx7587]

x=0

[@mircoceccarelli6689]

( 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 😊🤪👍👋

[@jumpjump-oz2pr]

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

[@googlem7]

multiplicity solution at end has been repeated

[@ИринаРзаева-ф2с]

Ответ один, а и 0 тоже...

[@aurochrok634]

septic… hm… 😂

[@sonicbluster3360]

0

[@bobajaj4224]

My Ex was septic..

[@BRYANCHONGYOUCHIANMoe]

can you answer my questions i send to your email 🙏🙏🙏🙏it's secondary school problem and really need your help. Thanks

[@sarahlo5084]

Medical person me reads “septic” 🤒

[@diegoretosanchez2129]

X=0 😂

[@SidneiMV]

7(7¹x⁶ + 7⁶x¹) + 21(7²x⁵ + 7⁵x²) + 35(7³x⁴ + 7⁴x³) = 0 (7¹x¹)(x⁵ + 7⁵) + 3(7²x²)(x³ + 7³) + 5(7³x³)(x + 7) = 0 x[(x⁵ + 7⁵) + 3(7x)(x³ + 7³) + 5(7²x²)(x + 7)] = 0 *x = 0* (x⁵ + 7⁵) + 3(7x)(x³ + 7³) + 5(7x)²(x + 7) = 0 x³ + 7³ = (x + 7)³ - 3(7x)(x + 7) x⁵ + 7⁵ = (x + 7)⁵ - 5(7x)(x³ + 7³) - 10(7x)²(x + 7) x + 7 = a 7x = b (x + 7)⁵ - 5(7x)(x³ + 7³) - 10(7x)²(x + 7) + 3(7x)[(x + 7)³ - 3(7x)(x + 7)] + 5(7x)²(x + 7) = 0 a⁵ - 5b(x³ + 7³) - 10ab² + 3b(a³ - 3ab) + 5ab² = 0 a⁵ - 5b(a³ - 3ab) - 10ab² + 3b(a³ - 3ab) + 5ab² = 0 a⁵ - 5a³b + 15ab² - 10ab² + 3a³b - 9ab² + 5ab² = 0 a⁵ - 2a³b + ab² = 0 a(a⁴ - 2a²b + b²) = 0 a(a² - b)² = 0 a = x + 7 = 0 => *x = -7* a² = b => (x + 7)² = 7x x² + 14x + 49 = 7x x² + 7x + 49 = 0 x = (-7 ± 7i√3)/2 *x = (7/2)(-1 ± i√3)*

[@Danish53879]

Mei muslman hon hindu nhi hon

[@Alfi-rp6il]

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.

[@PrimeNewtons]

I'll do that. There is that argument that every number is complex. What do you say? Also consider the argument that if a number has an imaginary part, it is altogether imaginary.

[@Alfi-rp6il]

@@PrimeNewtons Don't play tricks with words, ok. Mathematics is a science, not part of rhetorics.

[@PrimeNewtons]

You did not address my questions. It's no wordplay. You should at least say something about the validity of the claims. Let me repeat then here: 1. Every real number is a complex number with zero imaginary part. 2. If the imaginary part of a complex number is not 0, then it is an imaginary number. Not necessarily purely imaginary.

[@Alfi-rp6il]

@@PrimeNewtons No. Concerning 2.: A complex number is an imaginary number, when the imaginary part is not 0 and the real part IS ZERO.

[@PrimeNewtons]

I'm going to pose this question in the community. I need to learn more.

[@KramerEspinoza]

❤