End Result Will Blow Your Mind!

[@arulbiswas1260]

You could also use de'moivre's theorem. The roots are √3/2±i1/2 or z = {cos(π/6) +isin(π/6) }. We know, z^n = cosnx+isinx.

[@AniketKumar-lw6su]

Yea that's exactly how I did it too roots are W/i and its negative W³=1 and i³=-i.

[@AdityaYadav-hm1rh]

EXACTLY, bro just like to make it loooonngggg for no reason

[@markvarga9507]

Good one, I usualy do it like e^i*π/6, but it is the same thing

[@yashmehta9299]

Much simpler way to do this. x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x) = 0 So x^3 = -1/x^3 So x^999 = -1/x^999 by raising both sides to 333 So x^999 + 1/x^999 = 0

[@davidma16]

That complex solution and 999 power really screams Moivre’s formula

[@UttaraDhaka-v7t]

Ain't no way oxford has an entrance exam...💀

[@vaskaktm]

i think it does

[@Amit_Pirate]

The question is probably from 50 years ago

[@irenehartlmayr8369]

What are the entrance requirements for Oxford then ?...and not everyone wants,or needs,a degree in mathematics!

[@saturday9032]

Wait till you hear about tmua

[@saturday9032]

@@irenehartlmayr8369 you get a different question depending on your subject. You get a question once you do your interview at Oxford, it basically contributes to whether you can make it into the school. For economics for example you'd get a question regarding the economy, choices or something else. If you give a good answer you'd be considered for a place in Oxford so it's essentially like an entrance exam

[@fareednadeem4266]

This type of easy questions is asked in government exams in India 😂😂

[@arsh_arora10]

But still we Indians proudly lag behind in mathematics and innovation. It's not about difficulty

[@fareednadeem4266]

@arsh_arora10 yes because of the education system

[@TanmaySharma-o7y]

Also because of your mentality

[@fareednadeem4266]

Not my mentality but most of the indian students.

[@twinopedia7652]

well then try to solve JEE ADVANCED MATHEMATICS(NOT of 2024)

[@abacademy8896]

when you got x^3 = -1/x^3... can't you just power 333 both sides and get x^999 = -1/x^999 and then simply x^999 + 1/x^999 = 0... what is the problem if we do this?? bcz I don't feel any problem doing this

[@IamAlive-zk8jw]

doing power 333 both side make this like this: x^999=-1/x^999 x^999 + 1/x^999 = 0 so this is verified by two trees so this is correct, you found a new way

[@paulortega5317]

Let f(x,n) = xⁿ + 1/xⁿ, find f(x,999) f(x,n+2) = f(x,1)•f(x,n+1) - f(x,n) = √3•f(x,n+1) - f(x,n) f(x,0) = 2 f(x,1) = √3 f(x,2) = 1 f(x,3) = 0 and he sequence repeats every 12 terms So f(x,999) = f(x,3) = 0

[@brain_station_videos]

This is good. Btw how you figured out the second step? That recurrence relation..

[@paulortega5317]

@@brain_station_videos It was related to a problem posed by Sybermath. I began with the general form of it: f(x,a)•f(x,b) = f(x,a+b) + f(x,a-b). Rearranging it to f(x,a+b) = f(x,a)•f(x,b) - f(x,a-b). Now let a = n+1 and b = 1. The next step was to come with a formula for the nᵗʰ row coefficients for f(x,n) where for f(x,1) = c, in terms of powers of c. c c² - 2 c³ - 3c c⁴ - 4c² +2 c⁵ - 5c³ + 5c c⁶ - 6c⁴ + 9c² - 2 etc I figured out the formula and then discovered this was already known, per the On-Line Encyclopedia of Integer Sequences (OEIS). "A034807 Triangle T(n,k) of coefficients of Lucas (or Cardan) polynomials." Ha!

[@dsmithrus]

I just made a bruteforce. From the quadratic equation I got x1,2 = (√3±i)/2. Then, according to Moivre's theorem, x1,2 = 1*(cos(π/6) ± i*sin(π/6)) --> (x1,2)^999 = 1*(cos(999π/6) ± i*sin(999π/6)) = cos(π/2) ± i*sin(π/2)) = 0 ± i = ±i. Finally, by substitution, i+(1/i) = 0; (-i) + 1/(-i) = 0. The only answer is 0.

[@sanyalitvyak9571]

The complex exponential gives a hint that this can be generalised for x + 1/x = 2 cos(pi/2n) and x^n+1/x^n = 0. I think that the complex view gives quite a nice insight here.

[@penningmeestercgkdelft9159]

It is much quicker to use Euler's formula and note that the absolute value of x is just 1, whereas its argument is either +i*pi/6 or --i*pi/6. Multiply the arguments each by 999 and cancel redundant cycli of i*2*pi, from which it follows that x^999 = x^3 = 1 and x^-999 = x^-3 =-1, or the other way round. In both cases, hence for both roots of x, the sum is zero. Nothing to blow our minds, I suppose!

[@pizza8725]

Didn't expect to be this eazy

[@ezeonry]

you can turn that x value into an exponential and then divide 999 by 12 too man

[@abacademy8896]

also when we got x^3 + 1/x^3= 0, can't we just say that x*999 + 1/x^999 = (x^3)^333 + (1/x^3)^333... now as 333 is odd so it must has one factor= x^3 + (1/x)^3 = 0... Hence the whole expression becomes 0... so x^999 + 1/x^999 = 0

[@NamLe-dp1mx]

That’s what I thought too

[@yashmehta9299]

No, because (x^3 + 1/x^3)^333 is not necessarily the same as x^3*333 + 1/x^3*333

[@abacademy8896]

@@yashmehta9299 I am not doing this... we had got that x³ + 1/x³ = 0, so x³ =(-1/x³), now we can power 333 on both the sides and we will get- x⁹⁹⁹ =(-1/x⁹⁹⁹), now we will add 1/x⁹⁹⁹ on both the sides to get- x⁹⁹⁹ + 1/x⁹⁹⁹ = 0

[@yashmehta9299]

@@abacademy8896 ah, lol, yeah

[@omerbar7518]

This is literally just square both sides define u = x^2 solve for u find x. Put in original equation and see if it works. Done.

[@encounteringjack5699]

(e^(i[pi/6]))^999 Multiply pi/6 by 999, divide 2(pi), now having 333/4, we get 83.25, which means we now have 2(pi)/4, which is pi/2. Now we have e^(i(pi/2)) This will be at the bottom the fraction shown as 1/(x^999). Have e^(i(pi/2)) over 1 to form a fraction and multiply by e^(i(pi/2)) on top and bottom. The pi/2 and pi/2 will combine to make pi. Giving e^(i(pi)) which is -1. Denominators are the same and reduces down to zero. Same process can be used for e^(-i(pi/6)).

[@sriram.b5242]

Ahh yesss 1 + 1/1 is indeed equal to root of three

[@Self_improvement142]

😅😅

[@kroX-x1d]

bro pls upload more of there tricky questions more hard with good explanation ,,,,,[in same style]'''''''pls more '''thx

[@rajjhalani4354]

It's too easy I did it by another method:- x+1/x= √3 Squaring both sides,we get x²+1/x²=1 Solving this,we get x²=-w,-w² (Where w & w² are cube root of unity) Now putting any value of x² in x⁹⁹⁹+1/x⁹⁹⁹ will give answer

[@DavidGetling]

Try telling the truth. Unlike you, Oxford wouldn't embarrass itself by making a fuss over such SIMPLE quadratics.

[@vishalmishra3046]

*You were done mid-way (**1:43**) into the video* If x^3 = -1 / x^3 then x^999 = (x^3)^333 = (-1/x^3)^333 = -1/x^999. Therefore, x^999 + 1/x^999 = 0

[@Ashwinmathsclasses]

Very nice😮😮😮😮😮😮😮😮

[@DEVISRIPRASAD-n9p]

I solved it by substituting x (from the quadratic equation)in x^999 + 1/x^999 and by using de Moivre's theorem I got the answer zer0 . After I posted the comment and I saw you revealed another way also 😂😂

[@Changingcolors531]

How many ways the three balls can it be arranged when you get it from the group of six balls?

[@prashantgupta43905]

First, we choose 3 balls out of 6 in 6C3 ways and then arrange them in 3! ways So the ans will be 6C3×3! = 120 👍🏻

[@RahulChowdhury-y8r]

Nice☺☺☺☺☺☺☺

[@Changingcolors531]

How to calculate this "10!"?

[@Amit_Pirate]

n! = n(n-1)(n-2)(n-3)...(3)(2)(1) Therefore: 10! = 10×9×8×7×6×5×4×3×2×1

[@yashmehta9299]

Just say “10” loudly

[@ThePandey-q1k]

Bro we can manipulate root as cube root of unity 🫣

[@VIOLA4-D]

I don't know if my method is correct, I done it simply by differentiating separately and using log formula I got 0 as answer

[@Christopher-e7o]

X,2×+5=8

[@wanabecoderguy7]

make videos like ( or animation, to be more specific ) like 3blue1brown.

[@heenakhandelwal8608]

He has his own style, and I respect that and also love it. Why make everything so complex when this itself is super easy to follow

[@OsuAndChess]

And it's also the fact that if he does this he'd get hate for exactly that, reminds me of all "Mrbeast" type of videos

[@acidomegapro4893]

It can be changed to a cube root of unity so…. Yeah

[@kumkumsana409]

I am an eight standard student . I solved this problem within 2 min .😂

[@SANTHKAMALT]

Yes we can solve by euler formula🎉

[@MATH-_-MATICSOP]

here x is - of complex no. W so W⁹⁹⁹ =1 so ans is 2

[@x.in_hype]

Let . Step 1: Express Using the exponential form of , we have: x = e^{i\pi/6}. x^{999} = \left(e^{i\pi/6} ight)^{999} = e^{i \cdot 999 \cdot \pi / 6}. Simplify the exponent: 999 \cdot \frac{\pi}{6} = 166.5\pi = 166\pi + 0.5\pi. 166.5\pi \mod 2\pi = 0.5\pi. x^{999} = e^{i \cdot 0.5\pi} = i. --- Step 2: Find The reciprocal of is: \frac{1}{x^{999}} = \frac{1}{i} = -i. --- Step 3: Compute Now, substitute: x^{999} + \frac{1}{x^{999}} = i + (-i) = 0. --- Final Answer: x^{999} + \frac{1}{x^{999}} = 0.

[@gush5436]

chatgpt momento, haha ha h

[@brain_station_videos]

Good 👏

[@ManiTheObbyist]

def chatgpt because of the "\frac{1}{69}\" format on fractions

[@m9801]

Tried avoiding the value of x to no avail

[@maybeyourbee]

Ohhhh,my Head

[@ethancooper4154]

How did we manage to turn math into brainrot

[@ManojkantSamal]

Omg!The answer may be zero....... Based on the key x^6=(-1)...(^=read as to the power ) Explain later

[@IamAlive-zk8jw]

i am an indian(asian) so this is easy for me

[@rohitkaushik2311]

Hello bro I am also indian. 😁 Tumhara naam kya hai

[@rohitkaushik2311]

But I am in 10th grade. Aur mai Haryana se hu

[@IamAlive-zk8jw]

@@rohitkaushik2311 mai 9 grade me hu

[@Hello1-t1y]

Pls I can't handle sin, cos, tan, pi, and i

[@brain_station_videos]

And therefore i have solved it without using them

[@raghvendrasingh1289]

👍 x+1/x = √3 cubing both sides x^3+1/x^3+3x.1/x(x+1/x) = 3√3 x^3+1/x^3 = 0 x^3 = - 1/x^3 (x^3)^333 = - (1/x^3)^333 x^999 = - 1/x^999 x^999+1/x^999 = 0 other methods - (A) x^2 - √3 x+1 = 0 by quadratic formula x = (√3 +/- i)/2 x = cos π/6 +/- i sin π/6 x = e^(+/- iπ/6) x^999 = e^(+/- 333πi/2) = { e^(2πi) }^{+/- 83) e^(+/- πi/2) = +/- i because e^(2πi) = 1 in both case value of given expression is zero because reciprocal of +/- i is -/+ i (B) x^3+1/x^3 = 0 (x^3)^2 = - 1 x^3 = +/- i x^999 = (+/- i)^333 = +/- i because (+/- i)^332 = 1 as i^4 = 1 in both cases value of given expression is zero. (C) x^3+1/x^3 = 0 now x^999+1/x^999 = (x^3)^333+(1/x^3)^333 if we factorise it one factor will be x^3+1/x^3 hence answer is zero. (D) x = cos π/6 +/- i sin π/6 x^999 = cos { 83(2π)+π/2 } +/- i sin { 83(2π)+π/2 } = +/- i x^999+1/x^999 = 0 (E) by quadratic formula x = (√3 +/- i)/2 x = i(- 1 - i√3)/2 , - i( - 1+i√3)/2 x = iw^2 , - iw x^999 = i^999 , - i^999 because w^3 = 1 x^999 = i^3 , - i^3 because i^996 = (i^4)^249 = 1^249 = 1 x^999 = - i , i value of given expression = 0

[@brain_station_videos]

Amazing

[@AbdallahBOURAIMA-wh7et]

Please can someone explain me how x^6 can be negative?

[@Alphamatics1234]

Have u heard about complex number

[@Rishith198]

well it's complex.... if u know about complex numbers then i= sqrt(-1) if x=i then x^6= -1 as: i x i = -1 | -1 x i = -i | -i x i = 1 | 1 x i = i using this cyclicity of iota u will get the answer as -1

[@AbdallahBOURAIMA-wh7et]

@@Rishith198 thanks for the explanation though

[@AbdallahBOURAIMA-wh7et]

@@Alphamatics1234 I don’t know anything about them

[@sirajzama8080]

A simple one you got equation as x³=-1/x³ so multiply both sides by x³ you will get --> x⁶=-x³/x³. Cancel out x³ on RHS to get -1

[@arekkrolak6320]

First equation is quadratic so you can solve in memory. Then substitute in second

[@RealQinnMalloryu4]

{x+x ➖}+{1+1 ➖ }/{x+x ➖}={x^2+2}/x^2=2x^2/x^2=2x^1 (x ➖ 2x+1).6 3^3 (x ➖ 3x+3). {x^999+x^999 ➖}+{1+1 ➖ }/{x^999+x^999➖ }={x^1998+2}/x^1998=2x^1998/x^1998= 2x^3^2^3^2^2^3/x^3^2^3^2^2^3 2x^1^1^1^1^1^1^1/x^1^1^1^1^1^1^1 2x^1/x^1 2x^1/ (x ➖ 2x+1).

[@vishwa7180]

Very easy

[@RiteshKumar-bp8dm]

Hmm ... this question was a little bit of no. Theory good good 👍🏻

[@blobthebobtheborb]

oh hell no

[@tingbrian5437]

just use de moivre