Math Conundrum

[pudley4]

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

[pudley4]

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

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>

[Jim Brier]

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.

[BruceZ]

x

[BruceZ]

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>