My formula
$$(z-y)(z^2+zy+y^2)=(y-x)(y^2+yx+x^2)$$ $$z>=y+2$$ $$y>=x+2$$ $$x>=2$$
I am trying to understand a geometric problem better by plugging my formula into Wolfram Alpha. Normally I get a result which helps me understand the issue more graphically, but this time it is only spitting out a single solution.
The solution from Wolfram Alpha written out below:
$$x>=2$$ $$y>=\frac{\sqrt[3]{x^3+\sqrt{x^6+96x^3+256}+48}}{\sqrt[3]{2}}+\frac{8\sqrt[3]{2}}{\sqrt[3]{x^3+\sqrt{x^6+96x^3+256}+48}}+2$$ $$z=(-1)^{2/3}\sqrt[3]{2y^3-x^3}$$
My questions
- Is this the only solution?
- How would I know or check this?
- Since z has a complex component (-1)^{2/3} in this solution it can only ever be a complex number, correct?
Anything that improves my understanding is greatly appreciated!
Your equation says $x^3 + z^3 = 2 y^3$. Note that you want $x,y,z$ to be all $\ge 2$. With $y = ((x^3 + z^3)/2)^{1/3}$, your inequalities are equivalent to $$ \eqalign{\frac{x^3}{2} + \frac{z^3}{2} &\le (z-2)^3\cr (x+2)^3 &\le \frac{x^3}{2} + \frac{z^3}{2}\cr x \ge 2\cr}$$ Now the resultant of the polynomials $(x^3 + z^3)/2 - (z-2)^3$ and $(x+2)^3 - (x^3+z^3)/2$ with respect to $x$ is $-108 (3 z^2 - 6 z + 4) (z-2)^3$ whose only real root is $2$. Thus in the region $z > 2$, the boundaries of the regions represented by the first and second inequalities don't intersect. We can then check that the second inequality is implied by the first in this region. So the answer can be written as
$$ \eqalign{y &= ((x^3 + z^3)/2)^{1/3}\cr 2 \le x &\le (2(z-2)^3 - z^3)^{1/3} \cr}$$
Note finally that $2 \le (2(z-2)^3 - z^3)^{1/3}$ if and only if $z$ is greater than the real root of $4 + z^2/2 = (z-2)^3$, which is $$ (28 + 4 \sqrt{17})^{1/3} + \frac{8}{(28+4 \sqrt{17})^{1/3}} + 4$$