Question

If `x^2 + x +1=0` find the value of `(x+1/x)^3 + (x^2+1/x^2)^3...(x^100+1/x^100)^3`?

Answer 1

`197`

Explanation:

`(a+b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3`
`=> (x^n + 1/x^n)^3 = x^(3n) + 3 x^n + 3x^-n + x^(-3n)`
`x^3 - 1 = (x-1)(x^2+x+1) = (x-1)*0 = 0 => x^3 = 1`
`=> x^(3n) + x^(-3n) = 1^n + 1/(1^n) = 2`
`=> (x^n+1/(x^n))^3 = 2 + 3 (x^n + x^-n)`
`"sum from n=1 till n=100 : "`
`sum = 200 + 3 [ (x^101 - 1)/(x-1) + (x^-101 - 1)/(1/x-1) ] - 6`
`= 194 + 3 [ x^101 - 1 + x - x^-100 ]/(x-1)`
`= 197 + 3 [ x^101 - x^-100 ]/(x-1)`
`= 197 + 3 [ (x^3)^33 * x^2 - (x^-3)^34 * x^2 ]/(x-1)`
`= 197 + 3 [ x^2 - x^2 ]/(x-1)`
`= 197 + 0`
`= 197`

`"With complex numbers, one can proceed as follows."`
`x^3 = 1 => x = "cuberoot(1)" = cos(pi/3) + i sin(pi/3)`
`"or " cos(2pi/3) + i sin(2pi/3).`
`"Take e.g. "x = cos(2pi/3) + i sin(2pi/3) = e^(2 pi i/3).`
`"Then "x^n + x^-n = 2 cos(2 n pi / 3)`
`"So we have"`
`(2 cos(120°))^3 + (2 cos(240°))^3 + (2 cos(360°))^3 +`
`+ (2 cos(120°))^3 + ... + (2 cos(120°))^3.`
`= -1 -1 + 8 - 1 - 1 + 8 - .... - 1`
`= 6*33 - 1`
`= 197`

Answer 2

`197`

Explanation:

If `x^2 + x +1=0`

find the value of

`(x+1/x)^3 + (x^2+1/x^2)^3+cdots+(x^100+1/x^100)^3`?

Solving `x^2 + x +1=0` for `x` we have

`x = 1/2(-1pm i sqrt3)= e^(pmi phi)`

with `phi =pm (2pi)/3` then

with `x = e^(i phi)`

`sum_(k=1)^n (x^k + 1/x^k)^3 = sum_(k=1)^n (e^(i k phi) + e^(-i k phi))^3 = sum_(k=1)^n(e^(3k i phi)+e^(-3k i phi)) +3sum_(k=1)^n(e^(i k phi)+e^(-i k phi))`

but

`sum_(k=1)^n(e^(3k i phi)+e^(-3k i phi)) = sum_(k=1)^n(e^(2k i pi)+e^(-2k i pi)) = 2n` and

`3sum_(k=1)^n(e^(i k phi)+e^(-i k phi)) = 3((e^(i(n+1)phi)-1)/(e^(i phi)-1)-1+(e^(-i(n+1)phi)-1)/(e^(-i phi)-1)-1)`

here with `n = 100, phi = (2pi)/3`

`(e^(i(n+1)phi)-1)/(e^(i phi)-1)+(e^(-i(n+1)phi)-1)/(e^(-i phi)-1) = -1`

then finally

`sum_(k=1)^100 (x^k + 1/x^k)^3 =200-3=197`

Answer 3

`197`

Explanation:

Given:

`x^2+x+1 = 0`

We want to evaluate:

`(x+1/x)^3+(x^2+1/x^2)^3+...+(x^100+1/x^100)^3`

Without loss of generality, let:

`x = omega = -1/2+sqrt(3)/2i`

The only other option is the complex conjugate `bar(omega) = omega^2 = 1/omega` and the result is going to be real valued anyway.

Note that:

`omega^3-1 = (omega-1)(omega^2+omega+1) = 0`

So:

`omega^3 = 1`

Let's look at the first few binomials:

`omega+1/omega = omega+bar(omega) = -1`

`omega^2+1/omega^2 = bar(omega)+omega = -1`

`omega^3+1/omega^3 = 1+1 = 2`

These three values repeat cyclically.

So the sum is expressible as:

`(-1-1+8)+(-1-1+8)+...+(-1-1+8)-1 = 33*6-1 = 197`