A proof to follow...
 

He's right, it is zero [img]/forums/images/icons/smile.gif[/img]

Re: A proof to follow...
 

The easiest way is to factor the equation:

x^3 + 8x^(-3) becomes

[x+2x^(-1)] * [x^2 - 2 + 4x^(-2)]

If you notice, the second factor can be found from the initial equation:

(x + 2/x)^2 = 6
[x^2 + 4 + 4x^(-2)] = 6
[x^2 - 2 + 4x^(-2)] = 0

Since this factor = 0, the equation equals 0

Re: Math Conundrum
 

Here's a complete solution which solves for x:

(x + 2/x)^2 = 6

x^2 + 4 + 4/x^2 = 6

x^4 + 4x^2 + 4 = 6x^2

x^4 - 2x^2 + 4 = 0

x^2 = [2 +/- sqrt(4 - 16)]/2 from quadratic formula

x^2 = [2 +/- j*sqrt(12)]/2 where j = sqrt(-1)

x^2 = 1 +/- j*sqrt(3)

x^2 = 2*exp(+/-j*pi/6) going to polar form

<font color="red">x = +/-sqrt(2)*exp(+/-j*pi/12) =
+/-sqrt(2)*[cos(pi/12) +/- j*sin(pi/12)</font color>

x is complex, not real or imaginary, and has 4 possible values

<font color="red">x^3 - 8/x^3 =
+/-sqrt(8)*exp(+/-j*pi/4) -/+ sqrt(8)*exp(+/-j*pi/4) = 0 </font color>

Answer
 

The answer is zero. Here is how I did it.

First of all, you should recognize that (x^3) + (8/x^3) is the sum of two cubes. Factoring the sum of two cubes results in: [x+(2/x)][(x^2)-(2)+(4/x^2)].

Now let us work out [x+(2/x)]^2 = 6.

This is [(x^2)+(4)+(4/x^2)] = 6. Then,

[(x^2)-(2)+(4/x^2)] = 0. But this is the second factor when we factored the sum of the cubes above. So, since the second factor is 0, the result must be 0.

Above post is mine *NM*
 

x

Correction
 

x^2 = 2*exp(+/-j*pi/<font color="blue">3</font color>) going to polar form

<font color="red">x = +/-sqrt(2)*exp(+/-j*pi/</font color><font color="blue">6</font color><font color="red">) =
+/-sqrt(2)*[cos(pi/6) +/- j*sin(pi/6)]</font color>

= +/-sqrt(6)/2 +/- j*sqrt(2)/2

in areement with Polarbear

x is complex, not real or imaginary, and has 4 possible values

<font color="red">x^3 - 8/x^3 =
+/-sqrt(8)*exp(+/-j*pi/2) -/+ sqrt(8)*exp(+/-j*pi/2) = 0</font color>