Find the real values $x,y,z$ such that $$\begin{cases} x+y^2+z^3=21\qquad (1)\\ y+z^2+x^3=71\qquad (2)\\ z+x^2+y^3=45\qquad (3) \end{cases}$$
Thank you everyone. This problem have some nice methods,
my idea $$(1)-(2),(2)-(3),(1)-(3)$$ But following is very ugly,
On
I calculated the groebner basis of the given polynomials as @user64494 mentioned. I used Maxima. The groebner basis can be calculated by
load(affine) grobner_basis([x+y^2+z^3-21,y+z^2+x^3-71,z+x^2+y^3-45],[x,y,z]);
or by the more performant
load(grobner) poly_reduced_grobner([x+y^2+z^3-21,y+z^2+x^3-71,z+x^2+y^3-45],[x,y,z]);
The first method calculates three polynomials of the following structure
$$ \bar{a} z +\sum_{i=0}^{26}a_i x^i$$ $$ \bar{b} y +\sum_{i=0}^{26}b_i x^i$$ $$ \sum_{i=0}^{27}c_i x^i$$
the last polynomial is a polynomial in $x$ and its greatest common divisor with its derivate is $1$ so the mutiplicity of each of its zeroes is $1$. That gives 27 different roots from the last equation and for each of this $x$ we can calculate exactly one $y$ and exactly one $z$ from the second and first equation. So the system has exactly 27 different solution triples. Only one of the has integer coeffizients. This can be easily verified by checking the factors of the constant term of the third polynomial. When solving the last polynomial numericaly all additional zeros were complex. But I did not estimate the errors of these results so I maybe there are some additional real solutions.
Here is the groebner basis:
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By substitution $x = -z^3-y^2+21$ we come to the system $$\{z^6+2y^2z^3+y^4+y^3-42z^3-42y^2+z+396=0,$$ $$-z^9-3y^2z^6-3y^4z^3-y^6+63z^6+126y^2z^3+63y^4-1323z^3-1323y^2+z^2+y+9190=0\} $$ in two variables $y,z.$ The necessary and sufficient condition for $z$ to be a root of the reduced system is its resultant in $y$ equals zero. One can read the theory or/and consider a simple example of the system $\{xy-3=0, x+y-4=0\}$ having $y^2-4y+3$ as the resultant in $x$. Computing the resultant of the reduced system in $y$ with help of Maple, we obtain $$z^{27}-189 z^{24}+15869 z^{21}-270 z^{19}-770589 z^{18}-806 z^{17}+ $$ $$2751 z^{16}+23703246 z^{15}+82077 z^{14}-1652484 z^{13}-476609381 z^{12}- $$ $$ 3301322 z^{11}+43400763 z^{10}+6247199406 z^9+64051684 z^8-$$ $$ 614744566 z^7-51522303964 z^6-586660519 z^5+4368480127 z^4+ $$ $$244239132451z^3+2045234869z^2-12927999002z-506350844104 . $$ Its integer zeros may be only the divisors of $506350844104=2^3\cdot7^3\cdot{22717}\cdot{8123}$. With help of Maple it is easy to determine that $z=2$ is the only integer root of the discriminant. Substituting it in the reduced system, we obtain $ \{y^4+y^3-26*y^2+126=0, -y^6+39*y^4-507*y^2+y+2130=0\}$. Let us continue to find integer solutions. Factoring $126=2\cdot 3^2\cdot 7$ and $2130=2\cdot3\cdot5\cdot71$ and making substitutions in the last system, we find $y=3$ is its unique integer root. At last, $x=21-3^2-2^3=4.$ See the same result and other solutions obtained with Maple in Maple workshheet exported as a PDF file of size 4.7 MB. I think the usage of Groebner basis methods to this end is the same, but in other formulas.