[@ProfessorWumbo]

I found that when we cube both sides, we get the original equation back. x^3-x^(-3)=x-x^-1. And 2187=3^7. (:

[@Matroskra]

Could you please elaborate? If you cube both sides you get x^3-x^-3-3x+3x^-1=-2sqrt{2}i . I'm not following what you mean.

[@ProfessorWumbo]

@@Matroskra Certainly! The equation can then be rewritten as x^3-x^(-3)-3x+3x^(-1) = x^3-x^(-3)-3(x-1/x). From the original equation we know that x-1/x=isqrt(2). So we can substitute it in. x^3-x^(-3)-3(isqrt(2))=-2(isqrt(2)). Adding 3isqrt(2) on both sides gives us x^3-x^(-3)=isqrt(2)=x-1/x

[@Matroskra]

@@ProfessorWumbo OH, amazing! Thank you so much hehe

[@maxhagenauer24]

What do you mean x^3 - 1/x^3 = x - 1/x? Where did that equation come from? If you cube the first equation you get x^3 - x + 1/x - 1/x^3 = -2sqrt(2i), you set it equal to 0 for some reason and brought half of it over.

[@ProfessorWumbo]

@@maxhagenauer24 my previous comment should explain it.

[@kavjay]

X - 1/X = sqrt(2)*i is also equivalent to sin(Y) = 1/sqrt(2), where X = exp(Y*i) , and thus, X = exp(pi*i/4) . This is the approach that I used to find the solution

[@cosmosapien597]

From x²=-1/x²,
if we multiply both sides by x², we get x⁴=-1,
and if we multiply both sides by x, we get x³=-1/x.
We don't need to do the squaring part and the indentity part.

[@Abhay0505]

That's soo awesome 👍❤

[@nikhilupscaspirant]

Teacher please make a video on unit digit and remainder
Such as... a^b .. if we divide this type of number ... what would be unit digit and remainder

[@PrimeNewtons]

Email me a problem

[@nikhilupscaspirant]

@@PrimeNewtons please send me your email

[@nikhilupscaspirant]

@@PrimeNewtons please send your email

[@JayRagon]

@@nikhilupscaspirant you should be able to find it on his youtube channel

[@NoobPerson-xp7nn]

There are two methods of doing this
1) Binomial Theorem
2) Modular arithmetic
The 2md one is quick and easy to understand

[@surendrakverma555]

Very good 👍

[@KPunktFurry]

i commented the whole video but vorget to sent it up WOW
so to make it short it was nice!
LG K.Furry

[@THRIFT-GAMING]

Teacher we can do it simply,
If x square is equal to minus 1 by x square then X raise to the power 2186-1/x raised to the power 2186
Then we have left x minus 1 by X
We have completed that x minus 1 whole square is equal to root 2 i square
When we will transpose squre that side than it will be root then x-1/x is
√2i

[@glorrin]

tell me if I am wrong
assume x^4 is 1
that means x^2 is 1 or -1,
so x^2 + 1/x^2 is either 2 or -2 so not 0
so x^4 cannot be 1 and must be -1
Therefore x^2 can be either i or -i
in both cases x^2 + 1/x^2 is 0
therefore x^2 - 1/x^2 is either 2i or -2i
From x^2 = i or -i, there are 4 solutions for x
testing all 4 only 1 solved x - 1/x = sqrt(2) i (I am most probably wrong here too)
x = (1/sqrt(2))(1-i)
then x^2 is -i
therefore x^2 - 1/x^2 is -2i
solving x^n - 1/x^n for any n in N
if n mod(4) = o, x^n - 1/x^n = 0
if n mod(8) = 1 or 3, x^n - 1/x^n = sqrt(2) i
if n mod(4) = 2, x^n - 1/x^n = -2i
if n mod(8) = 5 or 7, x^n - 1/x^n = - sqrt(2) i

[@charlefranje1882]

How x²+1/x² =0 ? It's impossible for any value of x

[@phill3986]

😊😊😊👍👍👍

[@gnanadesikansenthilnathan6750]

Sir learning is a process sir . More than learning it's imagination. Let's not equate learning with living

[@renyxadarox]

x⁴=-1

[@makehimobsessedwithyou6412]

you are so clever

[@ibrahimmahdy3702]

x^4 =-1

[@charlefranje1882]

And how (x²-1/x²)² =-2 it's impossible any square of number is always positive

[@godQlol]

x - 1/x = (sqrt2)i
x^2 + 1/x^2 = 0
x^4 = -1
x^2 = i, 1/x^2 = -i
or
x^2 = -i, 1/x^2 = i
(x^2187 - 1/x^2187)(x - 1/x)
= x^2188 + 1/x^2188 - x^2186 - 1/x^2186
= -1 -1 - (-1)(x^2) - (-1)(1/x^2) = -2
Hence, x^2187 - 1/x^2187 = -2 / sqrt(2)i = sqrt(2)i

[@ibrahimmahdy3702]

-(2)^.5 i because x^4 =-1

[@arikbrock3623]

That's not correct . X^2187 = (X^2188)/X = 1/X . The answer is -(sqr2)i

[@ibrahimmahdy3702]

(2)^.5 i but x^4 =-1

[@rakeshsrivastava1122]

x^4=-1.

[@bhagyashrigadekar8618]

If you love math ,then you deserves this like button (BTW 1st comment)🗿
👇