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$\large \sqrt[3] {a} -\sqrt[3] {b}- \sqrt[3] {c} =16$

$\large \sqrt[4] {a} -\sqrt[4] {b} - \sqrt[4]{c}=8$

$\sqrt[6] a -\sqrt[6] b - \sqrt[6] c =4$

I was able to find one solution:

$\large (\sqrt [6]a, \sqrt [6]b, \sqrt [6] c)\rightarrow (x^2, -y^2, -z^2)$ respectively

$\large x^2+y^2+z^2=4$

$\large x^3+y^3+z^3=8$

$x^4+y^4+z^4=16$

Squaring the first equation,

$x^4+y^4+z^4+2[(xy)^2+(yz)^2+(xz)^2]=16$

$(xy)^2+(yz)^2+(xz)^2=0$

Two of the variables have to be equal to $0$. Without loss of generality, let $\large (y, z)=0$ This means that $\large x=2$ , so the solution $\large (a, b, c)$ is $\large (4^6, 0, 0)$

Is there more that one solution? It really feels like there should be more since there are all these high powers.

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I am weak on minus signs. Substituting letting $a=x^{12}$ and so on sounds like a good idea. But then we will get some minus signs on substituting. Of course the solution you found works, and the missing minus signs make no difference since $y=z=0$. –  André Nicolas Jul 1 '13 at 6:32
    
@AndréNicolas I used the minus signs in order to make the equations with x, y, z to have only plus signs, but I just realized that it would only make it so for thee equation with the squares. I need to revise this a little bit. –  Ovi Jul 1 '13 at 6:35
    
@Ovi If $$y^2=-\sqrt[6]{b}$$ how did you get $$y^3=-\sqrt[4]{b}$$ –  metacompactness Jul 1 '13 at 7:39
    
Where did you take this problem from? –  metacompactness Jul 1 '13 at 9:11
    
@metacompactness I realized that, but that can be fixed by making $\sqrt[3] a , \sqrt[3] b, \sqrt[3] c \rightarrow (x^4, -y^4, -z^4)$ . Then the third equation will be $x^4+y^4+z^4=16$ , and the first equation will be $x^2-y^2-z^2$ but when I square this I get a similar result and I can still make the substitution $x^4+y^4+z^4=16$ and be left with $0=yz-xy-xz$ . This not only allows for the solution that I already found, but since all the terms aren't positive anymore it also leaves room for $(y, z) \not = 0$ –  Ovi Jul 1 '13 at 17:01

2 Answers 2

up vote 1 down vote accepted

Let $x=\sqrt[12]{a}$, $y=\sqrt[12]{b}$ and $z=\sqrt[12]{c}$. The system becomes: \begin{equation} x^4-y^4-z^4=16~..................................(1)\\ x^3-y^3-z^3=8~.................................. .(2)\\ x^2-y^2-z^2=4~...................................(3) \end{equation} Squaring equation $(3)$ and adding it to equation $(1)$, we obtain: $$\left(x^2\right)^2-\left(y^2+z^2\right)x^2+y^2 z^2-16=0$$ whose discriminant is $\Delta=\left(y^2-z^2\right)^2+64$. The sum of its two solutions $x_1^2$ and $x_2^2$ is $x_1^2+x_2^2=y^2+z^2$.

Now, let us substitute $x_1$ and $x_2$ in equation $(3)$, we'll obtain: \begin{equation} x_1^2-y^2-z^2=4\\ x_2^2-y^2-z^2=4 \end{equation} Adding the last two equations gives us: $$y^2+z^2=-8$$ This last equation and equation $(3)$ gives us: $x^2=-4$ that is $x=\pm 2i$ Putting the values of $x$ in equation $(1)$, we obtain $$y^4+z^4=0~$$ which together with the equation $$\left(y^2+z^2\right)^2=(-8)^2$$ gives us: $y^2 z^2=32$

So we have two numbers $y^2$ and $z^2$ whose sum is $-8$ and whose product is $32$: they are the uolutions of the quadratic equation $u^2+8u+32=0$ so $$y=2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)$$ and $$y=2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right)$$ or vice versa. So the solution I obtained until now are: $$(x,y,z)=\left(2i,2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)\right)$$ $$(x,y,z)=\left(2i,2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right)\right)$$ $$(x,y,z)=\left(-2i,2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right)\right)$$ $$(x,y,z)=\left(-2i,2^{\frac{5}{4}}\exp\left(\frac{3i\pi}{8}\right),2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right)\right)$$

I run out of time; I didn't use equation $(2)$ yet to find other solutions.

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In the case of complex $y$ and $z$ the discriminant $\Delta$ is not necessarily positive. –  user64494 Jul 1 '13 at 9:16
    
@user64494 You're right. –  metacompactness Jul 1 '13 at 9:20
    
@ metacompacthess : Also $$(x,y,z)=\left(2i,2^{\frac{5}{4}}\exp\left(\frac{5i\pi}{8}\right),2^{\frac{5}{4‌​}}\exp\left(\frac{3i\pi}{8}\right)\right) $$ does not satisfy the equation $x^3-y^3-z^3 = 8$. –  user64494 Jul 1 '13 at 9:25

Maple also produces 6 complex solutions $\{x = -.8435457951+1.254365878i, y = 1.285003263-1.910820082i, z = -.7543658777-1.852277550i\}, \{x = -.8435457951+1.254365878i, y = -.7543658778-1.852277551i, z = 1.285003263-1.910820081i\}, \{x = -1.312908410, y = .9693626140+1.441457468i, z = .969362615-1.441457468i\}, \{x = -1.312908410, y = .9693626140-1.441457468i, z = .969362615+1.441457468i\}, \{x = -.8435457951-1.254365878i, y = 1.285003263+1.910820082i, z = -.7543658777+1.852277550i\}, \{x = -.8435457951-1.254365878i, y = -.7543658775+1.852277551i, z = 1.285003263+1.910820081i\}$ of the system $\{x^2-y^2-z^2 = 4, x^3-y^3-z^3 = 8, x^4-y^4-z^4 = 16\}.$ It is too long for a comment.

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The solution is expressed in terms of the roots of the polynomial $2t^3+6t^2+9t+6.$ –  user64494 Jul 1 '13 at 8:36
    
How is that? The polynomial is third degree so it only has three roots. –  Ovi Jul 1 '13 at 17:35

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