solve { ((x^2 −xy+y^2 =16)),((y^2 −yz+z^2 =25)),((z^2 −zx+x^2 =49)) :}


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Question Number 184607 by mr W last updated on 09/Jan/23
$solve { ((x^2 −xy+y^2 =16)),((y^2 −yz+z^2 =25)),((z^2 −zx+x^2 =49)) :}$
$s o l v e$ ${\begin{cases} x^{2} - x y + y^{2} = 16 \\ y^{2} - y z + z^{2} = 25 \\ z^{2} - z x + x^{2} = 49 \end{cases}$
Commented by Frix last updated on 09/Jan/23

$Exact solution possible but not useful . . .$
Commented by MJS_new last updated on 11/Jan/23
$the system { ((x^2 +xy+y^2 =a^2 )),((y^2 +yz+z^2 =b^2 )),((z^2 +zx+x^2 =c^2 )) :} can be solved like this: 1. solve (2)−(3) for z 2. insert in (1) and (2) 3. let y=px and solve (1) for x^2 4. insert in (2) which leads to p^2 +αp+β=0 but the system { ((x^2 −xy+y^2 =a^2 )),((y^2 −yz+z^2 =b^2 )),((z^2 −zx+x^2 =c^2 )) :} leads to p^4 +αp^3 +βp^2 +γp+δ=0 which can only be exactly solved in some cases. with a=4, b=5, c=7 it′s not possible.$
$the system {\begin{cases} x^{2} + x y + y^{2} = a^{2} \\ y^{2} + y z + z^{2} = b^{2} \\ z^{2} + z x + x^{2} = c^{2} \end{cases} can be solved$ $like this :$ $1 . solve (2) - (3) for z$ $2 . insert in (1) and (2)$ $3 . let y = p x and solve (1) for x^{2}$ $4 . insert in (2) which leads to p^{2} + α p + β = 0$ $but the system {\begin{cases} x^{2} - x y + y^{2} = a^{2} \\ y^{2} - y z + z^{2} = b^{2} \\ z^{2} - z x + x^{2} = c^{2} \end{cases} leads to$ $p^{4} + α p^{3} + β p^{2} + γ p + δ = 0 which can only be$ $exactly solved in some cases .$ $with a = 4, b = 5, c = 7 {it}^{'} s not possible .$
Commented by mr W last updated on 11/Jan/23

$t h a n k s s i r!$
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