Question

If `x+1/x=-1` then What is the value of `x^247+1/x^187=`?

Answer 1

`-1`

Explanation:

As already obtained in the first answer,

`x=-1/2+-i sqrt 3/2=(r (cos theta+i sin theta))=re^(i theta), respectively`

r = 1.

For the first root, cosine is negative and sine is positive. `theta` is in

the second quadrant and is `(2pi)/3`.

For the second, both are negative. `theta` is in the 3rd quadrant.

So, it is `(4pi)/3`

Now, `x = e^((2pi)/3i) and e^((4pi)/3i)`

Accordingly,

`x^247+1/x^247`

`= e^(247(2pi)/3i)+e^(-247(2pi)/3i)` and

`e^(247(4pi)/3i)+e^(-247(4pi)/3i)`

`=2 cos (247(2pi)/3i) and 2 cos (247(4pi)/3i)`

`=2 cos (164pi+(2pi)/3) and 2 cos (329pi+pi/3)`

=2 cos (even `pi + (2pi)/3`) and 2 cos (odd `pi+pi/3`)

`=2 cos ((2pi)/3) and -2 cos (pi/3)`

`=2(-1/2) and (-2)(1/2)`

Both are the same -1...

Answer 2

`x^{247} +1/(x^{187}) = -1`

Explanation:

In the equation `x+1/x=-1` occurs `abs x=1` because making `x = 1+delta` and substituting there are not real solutions for `delta` in the equation `1+delta + 1/(1+delta)=-1`: then we can make `x = e^{i theta}`.

Substituting we have

`e^{i theta} + e^{- i theta} = -1`

but

`(e^{i theta} + e^{- i theta} )/2 = cos(theta)`

so the equation reduces to

`2 cos(theta)=-1`. Solving for `theta` we have:

`theta = pm 2/3 pi+2k pi` with `k = 0, 1,2,...`

Now `x^{247} = e^{pm i times 247 times2/3 pi} = e^{pm i2/3 pi}`

and

`x^{187} = e^{pm i times 187 times 2/3 pi} = e^{pm i2/3 pi}`

Putting all together

`x^{247} +1/(x^{187}) = 2 cos(pm 2/3 pi) = 2(-1/2)=-1`

Answer 3

-1

Explanation:

For lower class student I do it as follows

Given

`x+1/x=-1=>x^2+x+1=0`

So

`x^3=x^3-1+1=(x-1)(x^2+x+1)-1=(x-1)xx0-1=-1`

Now

`x^247+1/x^187=(x^3)^82 *x+1/((x^3)^62*x)=(1^82*x)+1/(1^62*x)=x+1/x=-1`

Will this do?