@@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
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.
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
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.
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
simply cube both sides. (x-1/x)3 = 2^3/2i. x3-1/x3 = 2^1/2i = x-1/x. the pattern repeats if you keep cubing both sides. and X^2187 = x^3 ^7 which is cubing both sides 7 times leading to same value of (2^1/2)i
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
Not for any value, it's impossible for any real number but there are 4 answers in the imaginary numbers x1 = sqrt( i ) x2 = sqrt( -i ) x3 = -sqrt( i ) x4 = -sqrt( -i )
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. (:
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.
@@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
@@ProfessorWumbo OH, amazing! Thank you so much hehe
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.
@@maxhagenauer24 my previous comment should explain it.
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
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.
That's soo awesome 👍❤
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
Email me a problem
@@PrimeNewtons please send me your email
@@PrimeNewtons please send your email
@@Nikhilchauhan-c2s you should be able to find it on his youtube channel
There are two methods of doing this
1) Binomial Theorem
2) Modular arithmetic
The 2md one is quick and easy to understand
i commented the whole video but vorget to sent it up WOW
so to make it short it was nice!
LG K.Furry
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
simply cube both sides. (x-1/x)3 = 2^3/2i. x3-1/x3 = 2^1/2i = x-1/x. the pattern repeats if you keep cubing both sides. and X^2187 = x^3 ^7 which is cubing both sides 7 times leading to same value of (2^1/2)i
Very good 👍
Sir learning is a process sir . More than learning it's imagination. Let's not equate learning with living
😊😊😊👍👍👍
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
How x²+1/x² =0 ? It's impossible for any value of x
Not for any value, it's impossible for any real number but there are 4 answers in the imaginary numbers
x1 = sqrt( i )
x2 = sqrt( -i )
x3 = -sqrt( i )
x4 = -sqrt( -i )
you are so clever
And how (x²-1/x²)² =-2 it's impossible any square of number is always positive
x^4 =-1
That's not correct . X^2187 = (X^2188)/X = 1/X . The answer is -(sqr2)i
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
x⁴=-1
If you love math ,then you deserves this like button (BTW 1st comment)🗿
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x^2+1/x^2=0; (x^2+1/x^2)^2=0; x^4+2+1/x^4=0; x^4+1/x^4=-2
-(2)^.5 i because x^4 =-1
x - 1/x = i√2
x² + 1/x² - 2 = -2
x² + 1/x² = 0
(x² + 1/x²)(x - 1/x) = 0
(x³ - 1/x³) - (x - 1/x) = 0
x³ - 1/x³ = x - 1/x
x⁶ + 1/x⁶ - 2 = x² + 1/x² - 2
x⁶ + 1/x⁶ = 0
(x⁶ + 1/x⁶)(x³ - 1/x³) = 0
x⁹ - 1/x⁹ - (x³ - 1/x³) = 0
x⁹ - 1/x⁹ = x³ - 1/x³ = x - 1/x
x¹⁸ + 1/x¹⁸ - 2 = x² + 1/x² - 2
x¹⁸ + 1/x¹⁸ = 0
(x¹⁸ + 1/x¹⁸)(x⁹ - 1/x⁹) = 0
x²⁷ - 1/x²⁷ - (x⁹ - 1/x⁹) = 0
x²⁷ - 1/x²⁷ = x⁹ - 1/x⁹ = x³ - 1/x³ = x - 1/x
.....
x^(3ⁿ) - 1/x^(3ⁿ) = x - 1/x
2187 = 3⁷
=> *x²¹⁸⁷ - 1/x²¹⁸⁷ = x - 1/x = i√2*
(2)^.5 i but x^4 =-1
x=cosa+isina
x^4=-1.
*x⁴ = -1*
*x⁴ ≠ 1*
final answer - luckily - is correct BUT there's a huge mistake (x⁴ = 1) in the way to the solution of the problem.