Don't panic. There are two basic strategies to solve questions like this
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- Опубліковано 28 чер 2024
- Questions like these are common on the JEE Mains, which is one of the standardized tests in India. Thanks to Sounak for the suggestion! I share 2 problems related to a known value for x + 1/x.
0:00 questions
0:51 solution 1
2:10 solution 2a
5:36 solution 2b
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Thanks!
Its class 9 question in india i already solved these type question
For olympiad practice 😂
Bro, it's class 8 Bangladeshi general question, too 😅@@Antriksh-nx1bug
@@shafin3365 As a person who is aware of both the Indian and bangladesi syllabus. Definitely it is not a part of the Bangladesi Syllabus. We have step below questions like this in CBSE and half a step below in ICSE of 9th standard.
@@Souparno_BiswasAs a Indian who is in cbse board these types of questions come in class 10th maths chapter 4 quadratic equation
Just take 2 tests
they do :)
Yes we do and the combined score of those 2 tests is 50% hardly 😂
seems your score is 0%
real
'A person who thinks'
You're INSANE if you think the average score is 50%. That's what the 2-3% people score in JEE Mains.
@@boredlife 50th percentile is median not mean
No it's mean here in India 😢
~14% students according to my calculation from my shift, 27 january.
@@JEE-oq1me lol, my fkd up shift😭😭
@@JEE-oq1methat's the easiest shift we had so far
That second problem is confusing. You didn't extend the x > 0 requirement for it as it is obvious that the equation has no real roots for x. Therefore it makes sense to convert it to polar form of a complex number.
The x>0 only applies to the first question
In the second one the only requirement would be x != 0
Either I screwed up or got a different answer
@@pog16384wtf lol
you can't compare real numbers with complex no... that's why including x>0 for question 2 doesn't make any sense at all
Use
Complex number
X=cosa+i sina
1/x=cosa-isinb
X+ 1/x = √3
=>2Cosa=√3
Cosa=√3/2
a=π/6
X^100 + 1/x^100= 2 cos 100.π/6
=> - 2 cosπ/3
=>-2(1/2)
=> -1 ##
Fastest solution thanks 🙏🙏🙏
Wow not every aspirant is maths Major.😂
@@Smoked.X yes bro,
But this is the fastest way to solve this problem
@@Hrishi02005 I regret not studying maths seriously from 10th standard and onwards.
@@Smoked.X but still you can also study higher mathematics by your own self
The problem is a bit confusing when "x>0" is somehow "carried over" to the second question. However, if x>0 is a constraint, the 2nd problem has no answer at all.
Love the complex answer. You could also look at e^(2iπ/3) and e^(-2iπ/3) and look at them on the complex graph: they have real co-ordinates -1/2 from basic trigonometry, and the complex parts cancel (conjugates), so the sum is -1.
These were one of the easiest questions. You should try JEE Advanced maths.
For the complex solution, just start with the polar form z=aexp(iθ), then use the exponential form of cosine : 2.cos(x) = exp(ix)+exp(-ix)
definitely prefer the complex numbers route. That just feels so much more familiar.
3rd way to solve problem 2: start with method 1 until you reach x^6=-1, then recognize that x must be an odd twelfth root of unity. A little investigation of the other equations tells you which pair of such roots it could be, then cancel the x^96’s and recognize that x^4 will have real part equal to -1/2, and like all roots of unity, its multiplicative inverse is its complex conjugate, so the answer is -1.
Immediately knew we were gonna need complex numbers, since x⁶ is always positive in the set pf real numbers
At 4:11 you can save a little computation work by noting that 100 = 6*17-2; that works nicer than 100 = 6*16 + 4.
Awesome problems and solutions.Thank so much, professor.
In both cases I multiplied by x and got a quadratic equation, then I solved for x.
This x I inserted into the second part of the question. That worked well, I compared my found values to those from the video.
In the second question I had to deal with complex numbers, that was a little difficult, but also worked.
Nice and interesting video, I admire the elegant alternative solving method, I had no idea of.
Best greetings!
Yeah but u don't really solve it by simply plugging in values.
Coz usually these types of questions are asked in our Ntse(National Talent Search Examination) exam which is a national level competitive exam and in this we don't have a lot of time to solve each question
@@epikherolol8189
I know. I am glad, that I am not a student and that I am not going to have an exam.
Sometimes I watch the videos, pause them and try to find the solution. Just for fun, without time pressure.
Sometimes I am lucky and I figure it out, but sometimes I fail and then I am lucky, if I understand the explanation, as some questions I find really hard.
But it is always interesting for me, and I try to keep up.
Best greetings!
@@uwelinzbauer3973 yea in my free time I also solve without putting time limit tho
In case of the first question you could square the required value to get x+1/x + 2 which is 7 hence the answer is root 7
oh wait that's what this guy did too nvm lmaoo
Very clever, elegant, inspiring …….a full package!!!!!……… esp. the first solution👏👏👏👏👏👏
In SSC cgl exam we need to solve both questions under 45 second.😊
I solved directly by complex numbers and went through a tedious process of calculating successive degrees of x^n + 1/x^n, only to observe that values repeat over a period of 12 (and "anti-repeat" for a period of 6). I had no clue as to why this happens and your solution is so insightful! Bravo 🎉
brooooo i did exactly same thing in reverse order first i find x^n + 1/x^n for n =1,2,3,4... 16 on n = 16 i observed it's repeating for every 12th n than after like 5 min that complex number method came in mind T_T
Problem 2: very good method, for someone who has never heard of complex numbers...Otherwise, solve the quadratic equation x^2 - (root3).x +1=0 which has two complex solutions, cos(pi/6)+_i.sin(pi/6), then apply de Moivre's formula.
For first problem i just made form "x^2-5x+1=0 with x>0", thats would be very easy to solve
new coment, so great solves! i needed this
In a level maths we're just taught to multiply by x and solve the quadratic then expand out for the solution, we would usually only get real roots though, in further maths we would just do the complex method
nice
Thank you for the video. If x+1/x=a and f(k)=x^k+1/x^k, then f(k+1)=f(k)*a-f(k-1). Then I found a pattern in this sequence, from some point it repeats every 9 members. Thank you.
I think the repetition is every 12 terms.
If k = 1; f = sqrt(3) *
If k = 2, f = 1 **
If k = 3; f = 0 ***
If k = 4, f = -1 ****
If k = 5; f = -sqrt(3)
If k = 6, f = -2
If k = 7; f = -sqrt(3)
If k = 8, f = -1
If k = 9; f = 0
If k = 10, f = 1
If k = 11; f = sqrt(3)
If k = 12, f = 2
If k = 13; f = sqrt(3) *
If k = 14, f = 1 **
If k = 15; f = 0 ***
If k = 16, f = -1 ****
So f(100) = f (12x8 +4) = f(4) = -1
This is a typical, don't be afraid, just take the direct approach kind of problem. The complex number solution is super elegant.
Do you know in india these. Questions are given to children at age of 15
@@toxiceditzzzznah, question 1 is for 12-13 aged kids
@@Feng_Q my mistake in India too approx at this age only
@@toxiceditzzzz Kuch bhi bologe kya jee toh 17 ki age mein hota hai 😂
@@VBM375padhai to 15-16 se shuru na, 11th me hi complex nos.
Genius👍🏻
Another way to solve the first problem is to assume x = y²
Then √x + 1/√x = y + 1/y
And from the first equation , we have y² + 1/y² = 5
(y + 1/y)² - 2 = 5
( y + 1/y)² = 7
y + 1/y = √7
And as y² = x
y = √x
So we get our final answer as
√x + 1/√x = √7
No need to insert value as y
We in india solve these problems in 8th grade
@@unknownwarrior8269 I didn't ask you where you are from and in which grade you guys solve these kind of problems , I just stated an alternate method to solve the discussed question . It's cool that you guys solve these problems from a young age but isn't India's education system essentially the worst
@@twinkle_pie i agree
@@twinkle_pie wait they actually teach stuff in us schools? I thought you guys only had school sh**tings.... btw africa is not a country and oh yeah europe is a continent and no the earth isn't flat and yes your forehead measures 2 football fields... us education system is the biggest joke in the world. cope.
By solving the complex numbers in the second method if we multiply divide by iota it would become w/i (w - cube root of unity ) by using the and then we just substitute use a property of cube root of unity w² + w = -1 and it gets cancelled out and we get - 1 as the answer
For the first problem I found it easier to multiply both sides of the equation by x and then get x²-5x+1=0 and then solve for x to solve the problem.
I took the same path, but somewhere on my way I went into a trap, I think:
x² - 5x + 1 = 0
x² - 5x = -1
(x - 2.5)² = - 1 + 6.25
(x - 2.5)² = 5.25
x1 = 2.5 + sqr(5.25) ~ 4.7912
x2 = 2.5 - sqr(5.25) ~ 0.2087
So I get two positive values for x, which both satisfy the condition x>0, but only x1 is correct for the first equation. Where did I get it wrong? 🤔 What was your solution?
@@lupus.andron.exhaustus x2 is also correct for the equation isn't it ?
@@lupus.andron.exhaustus both of these are correct. You can see from the original equation, that the solution must be a number and then its reciprocal. 1/4.7912 = 0.2087.
Remember the original problem didn't ask you to find x, but if you work out the solution of what they want, you can use either value of x and get the same answer.
I think you are correct. those both would come to to square root of 7 in given formula.@@lupus.andron.exhaustus
@@Tiqerboy You're right. Must have been due to a lack of coffee. ;) Thanks!
I have started to realise that in mathematics, patience is the key because you never have obvious answers to problems like these. So the key is patience and simple but effective approach. We need get rid of the quick solving mentality to get better if we are not a master at concepts.
Great video!
You can solve it easily using the concept of cube root of unity. x is one of the 3 cube roots of unity.
For 2nd question solve quadratic equation, root is omega , omega to the power 100 + 1/ omega to the power 100
= omega to the power 100+ omega square to the power 100 , since omega and omega square are multiplicative inverse of each other
Omega to the power x = omega to the power x/3
Then equation becomes omega + omega square
Now wtk omega + omega square +1 =0
This implies that omega + omega square = -1
Average score is 10%. 50% is scored by top 3% people.
bro getting 300/300 in JEE Mains is not uncommon these days, 50% is a joke
Average NEET user be like:
For the second problem, the square of x+1/x can be takan 50 times in a row. İn the third one , its understood that -1 will always come
For the solution 2b, you can notice that e^(2i*Pi/3) + e^(-2i*Pi/3) = 2cos(2Pi/3)
That was inspiring
Second question was easily doable by Complex Variables (I did it in my head with correct answer)
Which solutions is smarter: 2a or 2b ?
the manipulations are so cool!
The first problem is very obvious and I solved within 15 seconds 😅. In India these problems are considered to be the giver of free marks😂
Bruh. Not just india, it's obviously a warmup for those who are unfamiliar with the process so they can better understand the second question
Yeah ,these were in 8 th standards
Stop being delusional. Every people in every country after standard 8 or atleast maximum standard 9 can do it, unless, they never paid attention to the class.
Amazing ❤
I’ll have to revisit this video a few times
I went the 5 * 5 * 2 * 2 route. Needless to say, yours was quicker.
It’s pretty obvious that z^2 - sqrt(3)z + 1 = 0, and both roots complex.
thus both roots take form re^(+-it), but product is 1 by Vietta’s Formula ==> r = 1 ==> z = exp(+-it) = cos(t) +_ i sin(t)
hence z + 1/z == z + z* = 2Re(z) = sqrt(3) ==> Re(z) = sqrt(3)/2 = cos(pi/6) ==> t = pi/6. Using De Moivre,
==> z^100 + z^-100 = 2Re(z^100) = 2cos(4pi/6) = -1
A 12th grade won't know this though
You got me!
Solved the second one with complex numbers. It was obvious way forward
can also solve it using cube roots of unity (in this case cube roots of -1). That way you can solve it much faster.
I could quickly work out problem 1 in my head, but sighed and quickly gave up on problem 2, without trying much. I hope that they grade on the curve, in India.
These type of problems in India are generally asked in competitive exams like SSC and other government exams.
Hmm that's means only intelligent people join the government and not the corrupt ones, right? Right?
Just considered x is a complex number so we can write it in the Polar formula
x = cos(z)+isin(z)
And his numerical companion is x'
x'=cos(z)-isin(z)
x'=1/(cos(z)+isin(z)) =1/x
So we can now that x plus 1/x is 2cos(z)
x +1/x = 2cos(z) =sqrt(3)
2cos(z)=sqrt(3)
cos(z)=sqrt(3)/2
So z = π/6 +2kπ
When k=0 » z=π/6
x¹⁰⁰=cos(100z)+isin(100z)
So x¹⁰⁰ +1/x¹⁰⁰ = 2cos(100z)
= 2cos(100π/6)
2cos(50π/3)=2cos(2π/3)=-1
We can also solve through de moivre
I put that second equation into desmos and, It seems impossible because the lines never touch (It does not have a solution) or maybe it's irrational. It's either around 2^.5 or 0.5 something. It also suggests that 1^.5 may not be 1. Also, I am talking about the value of x. But ((3+5^(.5))/2)^.5 is super duper close. It was pretty hard to find that.
The solution is complex, so it doesn't show up on Desmos.
perfect ❤.
the first que can have two ans if you use the square of the difference of the values instead of the sum of them
It is direct usage of cube roots of unity
Me being a grade 11th math student from India , solved this using complex numbers method and Euler form 🙃(second method).
I Loved your use of the word "Simplify"...
My biology brain really doubted the meaning of "simple" after that explanation !!! 😂
this is simple maths question . we usually solve these in school
Second solution is like: just find x from equation x+1/x=sqrt(3) and put the number in expression x^100+x^(-100)
for the 2nd problem i think no real value of x will satisfy the eq. as its domain is (♾,-2] U [2,♾)
Problem 2: special cases + increasing power
It took me around 10 to 20secs to solve these questions and im in grade 11
They are nowhere near to actual jee advanced questions(except for the easy one)
I wonder what the graph of all n values till 100 look like.
There is another way that is easier to think of though it's a bit more process. FIrst step, try to get an answer for x^10+1/x^10, then you need to calculate (x^8+1/x^8)*(x^2+1/x^2). In the process you need to calculate (x^4+1/x^4)*(x^2+1/x^2). Once you get x^10+1/x^10=1, then you can get the final answer is -1.
x^10 + 1/x^10 is my f(5). My method is clearly like yours, but I don't see how knowing x^10 + 1/x^10 gives you x^100 + 1/x^100.
@@gibbogle Sorry, my bad. I made a mistake (x¹⁰+1/x¹⁰)²=x¹⁰⁰+1/x¹⁰⁰+2...though it leads to x¹⁰⁰+1/x¹⁰⁰=-1, but it is wrong. I failed in this question though it can do (x¹⁰+1/x¹⁰)²=x²⁰+1/x²⁰+2, x²⁰+1/x²⁰=-1, (x²⁰+1/x²⁰)², (x⁴⁰+1/x⁴⁰)², (x⁸⁰+1/x⁸⁰)(x²⁰+1/x²⁰)=x¹⁰⁰+1/x¹⁰⁰+x⁶⁰+1/x⁶⁰...finally you may get the right answer but the process is too complex.
I doubt the ans for 2nd question is -1 because by using AM GM inequality you can prove that "t+1/t" will never have solution for t in range (-2,2)
Hey guys. The second one could have been solved much more easily using just another complex no. method. First, we know that one of the complex roots of unity (3rd root). i.e. omega, = minus1/2 + (root3)/2. When we observe the RHS. We see root 3. That is, two times (root3)/2. So after multiplying omega by (-i) (-root(-1)). We get (-i omega). We see that this is the root of the above equation (x + 1/x = root3). Hence again substituting (- i omega) in the req. value. we get omega^100 + omega^200. Using the concept of complex roots of unity, we can say that it is equal to omega + omega^2. And it's value is -1. And hence the answer. It may seem complicated when read here. But if done personally, it will be a question with a solution of just 3 lines. /
7:20 How did you convert the euler form in rectangular form??
For the second one, I got x=i^(1/3), which is the same thing, but when I calculated is out, I got ((i^4)^25)^(1/3) + ((i^4)^25)^(-1/3)=1+1=2. I'm guessing I messed up somewhere in breaking up the i^(100/3)
If S is any point on the side PQ of ∆PQR and S is joined to R,prove that PQ+QR>PS+SR
For the 2nd question: I actually did a very long method where I found x^4 + 1/x^4 = -1 and then continuously squared to get to x^64 + 1/x^64 and then I found x^36 + 1/x^36 = 2 so then, I multiplied those 2 equations to get x^100 + 1/x^100 + x^28 + 1/x^28 = -2 and then I found x^28 + 1/x^28 (it’s -1) and plugged it in to get x^100 + 1/x^100 -1 = -2. Therefore: x^100 + 1/x^100 = -1. A much less elegant solution than the given one, but hey! If it works, then it works
Well you probably lost a lot of time
The second approach really threw me off as I only understand like the first 4 line of the solution 😂
Nice questions. These types of questions are asked occasionally for the university entrance exams in my country.😉
which country
Ok narcissist
@@prasunbagdi6112Worst comment I've seen, hahaha
Just came up with another way to solve the eqn after 5:21 ...got the eqn (x⁴=x²-1) from 2nd and 3rd eqns,I assumed x²=-(w)² where omega or w = cube root of unity and as w⁴=w, we can easily manipulate it into finding (w)¹⁰⁰...
Also w+w²+1=0(where omega and omega² are complex cube root of unity...we got w+w²+1 by sum of roots= -b/a in x³=1)
Btw ur approach is also amazing and eular form makes it easy to understand 😊
Explain please how x^6=-1, is it positive everytime?
x is a complex number in the second example. I think the presenter should have made that clear.
@@Tiqerboy thank you
I have a similar problem from my maths teacher:
If x+1/x is a whole number, prove that x^n+1/x^n is also a whole number (x is real and n is a positive integer).
I actually feel this is quite easy for jee mains.
Those questions were pretty simple , most students have already done many variants of that question in india thus jee uaually doesnt ask these questions
For the see the first one 0:06
let x = (sqrt x)^2 (eg 4= (sqrt 4)^2 = 2^2 =4)
Hence
1/ (sqrt x)^2 + (sqrt x)^2 =5 equation 1
Let 1/sqrt x + sqrt x = n
(1/sqrt x + sqrt x)^2 = n^2 square both side
1/(sqrtx)^2 + (sqrt x)^2 +2 = n^2
5 + 2 = n^2 (substitute the value for equation 1)
7 = n^2
n= sqrt 7 easy problem
Abhas Saini student here, and I can already say all these questions you have put here are v easy.
I kinda confused from the 2nd problem, I know that if for example x raised to an even power than the number must be positive, how is the x^6 is equal to -1?
I would love an explanation. It looks fun
I think that’s only for the real numbers. i^2 = -1. Sorry if I’m wrong
@@jeffthevomitguy1178absolutely correct. As soon as we see x^(even integer) we immediately know that we are off the real number line. The cunning thing to notice is that if we then assume complex numbers, nearly all of our real algebra transfers over so we simply think: "hmm, complex" and carry on regardless
JEE mains is the Qualifier here.
Try JEE advenced.
The solutions involve x = e^(±iπ/6) for the second one.
Whats the point of calculating the complex roots? You get the same theta by using the fact z+1/z = 2cos(theta)
Wow!
Interesting. I did this by a completely different method.
First I squared x + 1/x and showed that x^2 + 1/x^2 = 1
Then by repeatedly multiplying by x^2 + 1/x^2,
(e.g. (x^2 + 1/x^2)(x^2 + 1/x^2) = 1*1 = 1 = x^4 + 1/x^4 + 2, therefore x^4 + 1/x^4 = 1 - 2 = -1
then (x^4 + 1/x^4)(x^2 + 1/x^2) = -1*1 = -1 = x^6 + 1/x^6 + x^2 + 1/x^2 = x^6 + 1/x^6 + 1, therefore x^6 + 1/x^6 = -1 - 1 = -2 etc.)
and setting f(n) = x^2n + 1/x^2n I got
f(1) = 1
f(2) = -1
f(3) = -2
f(4) = -1
f(5) = 1
f(6) = 2
f(7) = 1
then it is clear that the cycle repeats, i.e. f(n+6) = f(n)
x^100 + 1/x^100 = f(50) = f(44) = f(38) = f(32) = f(26) = f(20) = f(14) = f(8) = f(2) = -1
or, more directly, since 50 = 8*6 + 2, f(50) = f(2).
Once again, I got the right answer by the wrong method. It was fun though.
In 2nd question every even power will have the answer = -1 bcz the pattern that follows is Y²-2 and Y = -1
In the second problem the answer is long since x^6 cannot be equal to -1 since no number multiplied by its self 6 times equals a negative number therefor the answer to problem 2 is no solution (impossible)
Pretty easy compared to the international olympiad
Wow just like that
It’s this sort of algebraic manipulation that I was never good at…I always took problems too literally and tried to solve for x all the time, endlessly confusing and frustrating myself. Honestly, I still struggle with this type of mind-bending 😁
Same bruh all Indians are commenting "oh ez took 3 secs"
Bro I am here trying the value of x for past 15 mins💀
z + 1/z = V3 ERROR
To understand what values the expression x + 1/x takes, we need to perform a proof. The given equation shows that x cannot be zero (0) because it is divided by zero. However, for the value x=1, the result of the equation will be the number 2. For the remaining values of x, we can assume that x = a/b, so 1/x will be b/a. this way we will obtain the equation that x + 1/x = a/b + b/a. Now let's solve the equation a/b + b/a. It looks like:
a/b + b/a = ( a^2 + b^2 ) / ab As you can easily see, from the given expression we can use a right triangle for analysis, where a^2 + b^2 give us c^2, which also means the area a square with side c, while "ab" is the result of the surface area of the rectangle which we will denote as S=ab. Therefore, we will obtain the formula that x + 1/x = c^2 / S. This means that we divide a square with side "c" by the area of a rectangle with sides "a, b". For these calculations, let's assume the individual values of these sides, using the Pythagorean theorem, i.e. a=3, b=4, c=5. Using this data, we obtain that c^2 = 25 and ab = 12. Now let's calculate c^2/ab = x + 1/x = 25/12 = 2+1/12.
And this is proof that x + 1/x is equal to or greater than 2. However, for negative numbers it will be equal to or less than -2.
Mixing two different concepts of math is a nice way to explore.like in a right angle triangle there are 3 lenths base, height and perpendicular, if we take two in a fraction then we can rearrange them into 3p2 ways that is 3×2=6 ways and that's why there are 6 trigonometry functions are exist.(sorry for my bad English)
Wow!
There goes my hope of discovering a new trig function...
The good thing is, you don’t need to know this in the real world.
In India, competitive exam candidates are informed of the following basic concepts. Your case involving √3 is one of them.
When dealing with equations having complex numbers as roots, the first step is to check if any of the following three forms are present (where "p" can be positive or negative without making a difference):
|Form 1| When the equation has the form x^2 + px + p^2 = 0, it leads to a specific conclusion: x^3 = p^3.
|Form 2| When the equation has the form x^2 + √2px + p^2 = 0, it leads to a specific conclusion: x^4 = (-p^4).
|Form 3| When the equation has the form x^2 + √3px + p^2 = 0, it leads to a specific conclusion: x^6 = (-p^6).
An interesting fact follows:
In exams at this level, questions are asked based on the above three forms when complex roots are involved. This is because these are the most straightforward cases. Among these, |Form 1|, |Form 2|, and |Form 3| are associated with taking complex numbers at 60°, 45°, and 30° on the Argand Plane, respectively. These are essentially fractions of 180°, namely one-third, one-fourth, and one-sixth, which is why raising these mixed numbers to the power of 3, 4, and 6 respectively yields real numbers.
Thanks😐
Neat.
you can literally just multiply both sides by x and get a quadratic equation but ok
2b is the best way to solve under exam pressure
to be honest, the first one wasn't that hard, pretty basic. The second one took some time, but, also solvable if the thought process is correct.
Second method is most advance vs first one.
Here in india we are taught to remember the values of cube roots of unity for these types of problems... A typical JEE aspirant can do this without even lifting a pen..
That's the problem just fking remember
No I studied in Allen they deeply explain the concepts hardly any coaching institute makes their student remember this values.
@@gitarthabordoloi5913 i mean it's all good if we know how they got derived in the first place
@@r8dra in my coaching the derivation is given how it happened,why it happened etc.But I agree school students are mostly made to remember this value's without the logic.
In problem 2 if X is real then 1st condition x+1/x =root 3 is impossible.....
The first one is actually very basic and straightforward.
The second one was a bit tricky. Here's how I did it,
01.x+(1/x)=√3
02.x^2+(1/x^2)=1
03.x^3+(1/x^3)=0
04. By multiplying equation 02 and 03,
x^5+(1/x^5)=-√3
05.x^10+(1/x^10)=1
06.x^20+(1/x^20)=-1
07.x^40+(1/x^40)=-1
08.Multiplying equation 06 & 07,
x^60+(1/x^60)=2
09.x^80+(1/x^80)=-1
10.Final step:Multiplying equation 09 & 06,
x^100+(1/x^100)=-1
I know it looks ridiculous but it works for me😅.
that's actually genius
@@kajaldey2656 Thanks😊
Ayoo you're a cuber too?