Math Conundrum
 

If the quantity [x + (2/x)]is squared and set equal to 6, what is the quantity [(x cubed) + ((8)/(x cubed))]?

A guess
 

zero

Re: Math Conundrum
 

Taking the square root of both sides and multiplying through by x, you can solve for x by the quadratic.
X=square root of 6 plus or minus square root of 14 all over 2. Plugging both results into your new equation, the result is approximately 29.39388 either way.

Re: Math Conundrum
 

180

Re: Math Conundrum
 

(x + 2/x)^2 = 6
(x + 2/x) = + - sqrt(6)
x^2 + 2 = + - (sqrt(6))*x
x^2 + - (sqrt(6))x + 2 = 0

x = ( + - sqrt(6) + - sqrt(6 MINUS 4*1*2) ) / 2

x = (+ - sqrt(6) + - sqrt(-2)) / 2

x is a complex number, however that doesn't mean the problem doesn't have an answer.

Re: Math Conundrum
 

The first step of the quadratic is -b, not +-b. It results in two possible answers for X, but they both give the same answer when plugged into the next equation.

Scratch that...
 

Oh, I see your point... I was assuming that we are only allowing the + square root of 6. Allowing for negative square root of 6 does make the problem a little more messy.

Re: Math Conundrum
 

No, if you use the quadratic equation to solve for X, you get an imaginary number, not a real number.

(x+2/x)^2=6
x + 2/x = sqrt(6)
x^2 - sqrt(6)x + 2 = 0

plug into the quadratic equation [-b +/- sqrt(b^2 - 4ac)]/2a

notice b^2 - 4ac = 6 - 8 = -2, so you end up with an imaginary number.

[edit] and I see this point has already been made [img]/forums/images/icons/smile.gif[/img] [/edit]

Re: Math Conundrum
 

Sorry, I missed the "is squared" in my first reply.
Imagine that!

solution
 

I am using the notations a^b = a to the b power, and sqrt(c) = the positive square root of c.

Expanding out (x + (2/x))^3, we get x^3 + 6x + 12/x + 8/(x^3).

Rearranging the terms,

(x + (2/x))^3 = x^3 + 8/(x^3) + 6(x + 2/x)

Now, we are given (x + 2/x)^2 = 6. Therefore (x + 2/x) = sqrt(6) or (x + 2/x) = - sqrt(6).

so, we have (sqrt(6))^3 = x^3 + 8/(x^3) + 6*sqrt(6), or
................(-sqrt(6))^3 = x^3 + 8/(x^3) - 6*sqrt(6)

solving each: 6*sqrt(6) = x^3 + 8/(x^3) + 6*sqrt(6)
0 = x^3 + 8/(x^3)

-6*sqrt(6) = x^3 + 8/(x^3) - 6*sqrt(6)
0 = x^3 + 8/(x^3)

In either case, x^3 + 8/(x^3) = 0, which is what we were asked to find. Looks like irchans is correct, it all disappears into nothingness.

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>