$\huge x^{1729}+x^{-1729}$

Find the value of the above expression if $x + \frac 1 x = 1$

The answer is 1.

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nice solution Rajdeep.....

varun patidar
- 6 years, 3 months ago

This is the best solution and this is easy for me to understand so thanks Rajdeep

Mohammad Raza
- 5 years, 7 months ago

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How come that 2cos(pi over 2) is 1?

Wendy Nicolas
- 5 years, 3 months ago

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Where have I done that ? I don't see $2\cos{(\frac{\pi}2)}$

Rajdeep Dhingra
- 5 years, 3 months ago

I thought I could do by looking at the pattern of.. Result of "n-1" for x^n + x^-n .. What's wrong wid dat?

jaswanth ayrus
- 5 years, 5 months ago

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You could do that too. @Kishlaya Jaiswal 's solution proves that in a little bit Complex way.

Rajdeep Dhingra
- 5 years, 5 months ago

Assuming y = x^{1729} + x^{-1729} Taking the 1729th root for the first equation will give as the second equation and then y=1

Haitham AlHazmy
- 5 years ago

The equation is the definition of infinite u can make it equal any number

Justin Allen
- 5 years, 6 months ago

But check your work, plug your solution back into the original problem, 1+1/1=2 not 1

Timothy Holmes
- 5 years ago

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They didn't say to find the value of x. S = 1 , not x.

Rajdeep Dhingra
- 4 years, 2 months ago

First, we strike to solve the given equation. Rearranging it, gives $x+\frac{1}{x} = 1 \Rightarrow x^2-x+1=0$ The roots of this equation are $x = \frac{1 \pm \sqrt{3}i}{2}$ The above value should immediately ring a bell if we remember that the $3^{rd}$ root(s) of unity are $\frac{-1 \pm \sqrt{3}i}{2}$ . Let us denote this by $\omega$ that is $\omega = \frac{-1 + \sqrt{3}i}{2} \qquad \text{and}$ $\omega^2 = \frac{-1 - \sqrt{3}i}{2}$ From this, we can immediately make out that $x = -\omega, -\omega^2$ .

But why did I represent $x$ in terms of $3^{rd}$ root of unity?

The answer is simple. Since we are dealing with a large power ( $1729$ ), expressing $x$ in terms of $3^{rd}$ root of unity can help us simplify because of the well-known properties : $1+\omega+\omega^2 = 0$ $\omega^3 = 1$

Also observe that $1728 \equiv 0 \pmod{3}$ .

And thus, we finally left with computing

$\begin{aligned} x^{1729} + x^{-1729} & = & (-\omega)^{1729}+(-\omega)^{-1729} \\ & = & (-\omega)^{1729}+(-\omega^2)^{1729} \\ & = & (-\omega)(-\omega)^{1728}+(-\omega^2)(-\omega^2)^{1728} \\ & = & (-\omega)+(-\omega^2) \\ & = & \boxed{1} \end{aligned}$

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Not posting this as a dispute since this a minor problem, but $[ x]$ square brackets usually denote the Greatest Integer function. Could you change the brackets to normal brackets? @shubhendra singh

Siddhartha Srivastava
- 6 years, 3 months ago

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Actually, I also took those brackets to be greatest integer function at first sight. But then, while solving I realized, it shouldn't be so.

And I can't change the problem statement because I am not a moderator and neither it's my own problem.

Kishlaya Jaiswal
- 6 years, 3 months ago

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I know. I didn't want to post a dispute as this seemed minor. Also, I had tagged the person whose problem this was but it seems I messed up and he wasn't tagged. Sorry for hijacking you solution. :(

Siddhartha Srivastava
- 6 years, 3 months ago

@Siddhartha Srivastava I'm really sorry if you got the problem wrong .......my mistake.

Shubhendra Singh
- 6 years, 3 months ago

I have edited it. Thanks.

Pranjal Jain
- 6 years, 3 months ago

This is an elegant way to solve it. Thanks for make me remember unit roots.

Adriana Lara
- 5 years, 3 months ago

X^1729 + X^(-1729) = X^1729 + (1/X)^1729 = [X + (1/X)]^1729 =1^1729 = 1 Ans.

Last Step Heaven
- 6 years, 3 months ago

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U cant take common!! man

Madin Ansari
- 6 years, 3 months ago

Hi!Sister.The method of solving above equation is may be wrong.Plz inform me what method you used to solve this equation.

মোঃ খান
- 5 years, 5 months ago

How did you conclude that -- X^1729 + (1/X)^1729 = [X + (1/X)]^1729

Kishlaya Jaiswal
- 6 years, 3 months ago

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oh sorry , mistake mistake , let X^1729 = x and X^-1729 = 1/x then X^1729 + X^-1729 = x + (1/x) = 1

Last Step Heaven
- 6 years, 3 months ago

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@Last Step Heaven – You're very wrong. Its like saying 5*10=100√5=50!!!!!!!

Prawar Pariyar
- 6 years, 3 months ago

yr wrong : as suppose x^2+x^-2 not equal to (x+1/x)^2

Niliansh Sinha
- 6 years, 3 months ago

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Amazing solution.

Aayush Patni
- 6 years, 3 months ago

awesome...

Himanshu Gupta
- 6 years, 3 months ago

Great man.....

varun patidar
- 6 years, 3 months ago

$\begin{aligned} x + \frac 1 x & = & 1 \\ x^2 + 1 & = & x \\ x^2 - x + 1 & = & 0 \\ (x+1)(x^2-x+1) & = & = 0 \\ x^3+ 1 & = & 0 \\ x^3 & = & -1 \\ \end{aligned}$

Knowing that $1729$ is a taxicab number, therefore $179 = 1^3 + 12^3 = 9^3 + 10^3$

$\Rightarrow x^{1729} = x^1 \cdot x^{12^3} = x \left ( (x^3)^4 \right)^{144}$

With $x^3 = -1 \Rightarrow x^{1729} = x \cdot \left ( (-1)^{144} \right ) = x$

$\Rightarrow x^{1729} + \frac 1 {x^{1729}} = x + \frac 1 x = 1$

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You nailed it man! This is the FACT that everybody is looking for... that 1729 is a special number!

Justin Augustine
- 6 years, 3 months ago

Most easy method dude. ....

Syed Faraz
- 5 years, 6 months ago

My favourite way of solving $x+\frac{1}{x}$ questions:

Let $x = e^{iy}$ ; then $\cos y = \frac{e^{iy}+e^{-iy}}{2} = \frac{1}{2}$ . One of the solutions for $y = \pi/3$ , but any other solution will work.

It follows that $x^{1729}+x^{-1729} = e^{1729iy}+e^{-1729iy} = 2\cos 1729y = 2\cos(1729\pi/3) = \boxed{1}$ .

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x^2+(1/x)^2=-1.

x^4+(1/x)^4=1.

...... Look for periodicity.
......
x^1726+(1/x)^1726=-1.

x^1728+(1/x)^1728=1. ......eq.1.

@ Multiply eq.1 by x once(eq.2) and divide by x (eq.3).add eq.2and 3(eq.4).

T+x^1727+(1/x)^1727=1. (T =x^ 1729+(1/x)^1729).

Repeat the procedure in @.

Putting the known values, T=1.

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If you have the same base and it's addition or subtraction you just need to add the exponent of the two base

1729+(-1729)=0

so the exponent is zero and we know that if the exponent is zero it is equal to one x to the power of 0= 1

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No that's what you multiply not add or subtract .

Khizer Khani
- 5 years, 10 months ago

The problem is very easy to solve. They have given , x + 1/x=1

=x + x^ -1= 1

=x^1 + x^ -1 =1 ......(1)

now if we follow the question, what we want is x ^1729 +x^-1729 to get 1729 and - 1729 as the power of bases of(1), I can multiply the power 1 and -1 with 1729 to get expected result: so even on the rhs, we multiply the power with 1729= 1^1729= 1.

I have not used the proper terms so plz excuse me.. whatever I have shown is the logic I used... I might be wrong, and I might be right, so please correct me if its the former case.

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That is not something you can do, it just happened to work in this case!

Ojas Singh Malhi
- 3 years, 9 months ago

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Given that x+1/x=1, So relating the above condition with x^1729+x^-1729 Gives 1 as the result.

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This does not only work for 1729, it works for all 6k + 1, where k is any integer.

Following Rajdeep Dhingra's solution, and replacing 1729 with 6k + 1, we have $S = 2\cos (\frac{(6k+1)\pi}{3}) = 2\cos(2k\pi + \frac{\pi}{3}) = 1$

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The equation at the bottom means x = 0.5 Cancel out the 2 powers and you have 0.5+0.5 = 1

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Let $a_{n} = x^n + x^{-n}$ . Then since $a_1 = 1$ by multiplying it out, $a_n = a_n.a_1 = a_{n+1} + a_{n-1}$ . This gives us $a_{n+1} = a_{n}- a_{n-1}$ .

Using this recurrence you see that the sequence of values has period 6, so $a_{1729} = a_1= 1$ .

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Nice! It is one of the most elementary solutions. Unlike some other solutions, it is general enough to solves the problem not only for 1729, but for any
*
n
*
. It is astonishing that this solution did not appear at the very beginning.

Xuefeng Wen
- 5 years ago

ln x^y + ln 1/x^y= ln z

y
*
ln x - y
*
ln x=0=ln z

z=1

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If we analize the series we get to know after finding two values (x+1/x=1) and (x^2 + 1/x^2= -1) Now the value of further series is difference of preciding two like (X^3 + 1/x^3 = -2) difference of values of (x^2 + 1/x^2) & (x^1 + x^1)

Doing the same for upto (x^6 + 1/x^6 = 2)

We now get that from x^7 the value will get repeat and it attains the value of same as of (x + 1/x) Which means the value x is changing as ( X^n + 1/x^n = X^(6k+m) )

Where n is the power of x can be written as 6k+m

Where k and m depends on the number i.e. x^13 K=2 & m=1

For m=1 it will attain the value of (x + 1/x =1) M=1. 1 M=2 -1 M=3 -2 M=4 -1 M=5. 1 M=6 2

now in given question N= 1729 Which can be written as 288*6+1 Here k= 288 and m= 1 So for m=1 value is 1

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Just make the given equation as per the another one.

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I know that many of you will laugh reading this but...

I simply solved it like that:

$x^{1729}+x^{-1729}$ with $x+\frac{1}{x}=1$

$x^{1729}+\frac{1}{x^{1729}}$

hence, [x=1]\

What's wrong in my reasonment? I'm a disaster at maths xD

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Simply both x powers are combine -1729+1729=0 ,and x°=1.....

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Since x+1/x= 1, then x^2 + 1 =x and 1+1/x^2 =1/x.

Then since (x+1/x)^2 = x^2+1/x^2+2, using identities derived above we further derive x^2+1/x^2+2 = x^2+1/x+1 = x + 1/x. Thus it also holds (x+1/x)^3 =1.

Since 1729 = 3^3 * 2^6+1.

Then It follows (x+1/x)^1729 = ((x+1/x)^3)^3 * ((x+1/x)^2)^6 *(x+1/x)^1 = 1, based on previous identities.

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This for me was quite easy to solve without using too much work. All you have to know is that there cannot be negative powers, and if there are, you make them denominators. Therefore you would have

x^{1729} over x^{1729}

Simplify To get 1

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Nice question given x+1/x=1 =x^2=x-1 Now x^4=(x-1)^2=x^2-2x+1=x-1-2x+1=-x

X^8=(x^4)^2=(-x)^2=x-1

X^9=x^8.x=(x-1).x=x^2-x=x-1-x=-1

X^81=(x^9)^9=(-1)^9=-1I

X^729=(x^81)^9=(-1)^9=-1

X^10=x^9.x=-x

X^100=(x^10)^10=(-x)^10=x^10=-x

X^1000=(x^100)^10=(-x)^10=x^10=-x

Now

X^1729+x^-1729 =x^729+1000+1/x^729+1000

=(x^729). (X^1000)+1/(x^729). (X^1000)

=(-1). (-x)+1/(-1). (-x)

=x+1/x

=1

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First, we experiment with

$x+\frac{1}{x}$

Square it. Since

$(x+\frac{1}{x})^2 = x^2+\frac{1}{x^2} + 2\frac{x}{x} = 1$ ,

and we know that $1^2$ is $1$ so it goes on and the answer is 1

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×

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First we solve the equation $x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0$ The roots of this equation are $x = \frac{1 \pm i\sqrt{3}}{2}$ This root can be written as $x = \cos{\frac{\pi}{3}} \pm i\sin{\frac{\pi}{3}}$ Now let $x = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}$ and $\frac{1}{x} = \cos{\frac{\pi}{3}} - i\sin{\frac{\pi}{3}}$

We need to find $x^{1729} + (\frac{1}{x})^{1729}$ . Let it be equal to $S$

Using de moivre's theorem( de moivres theorem ) we get:- $S = (\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}})^{1729} + ( \cos{\frac{\pi}{3}} -i\sin{\frac{\pi}{3}})^{1729}$

$S = 2 \cos{\frac{1729\pi}{3}}$

$S = 2 \cos{(576\pi + \frac{\pi}{3})}$

$S = 2 \cos{\frac{\pi}{3}}$

$S = 2 \times \frac{1}{2} = \boxed{1}$