x^3 +y^3 =z^2 x^3 +z^3 =y^2 y^3 +z^3 =x^2 obviously x=y=z=0∨(1/2) trying to totally solve it let y=px∧z=qx (p^3 +1)x^3 =q^2 x^2 (q^3 +1)x^3 =p^2 x^2 (p^3 +q^3 )x^3 =x^2 ⇒ x=0 ⇒ y=0∧z=0 ★ x=(q^2 /(p^3 +1)) x=(p^2 /(q^3 +1)) x=(1/(p^3 +q^3 )) ⇒ (p^2 /(q^3 +1))=(q^2 /(p^3 +1)) (p^2 /(q^3 +1))=(1/(p^3 +q^3 )) ========== p^5 +p^2 −q^5 −q^2 =0 p^5 +p^2 q^3 −q^3 −1=0 subtracting both p^2 (q^3 −1)+q^5 −q^3 +q^2 −1=0 (q−1)(p^2 (q^2 +q+1)+q^4 +q^3 +q+1)=0 ⇒ q=1 ⇒ p^5 +p^2 −2=0 (p−1)(p^4 +p^3 +2p+2)=0 ⇒ p=1 ⇒ x=y=z=(1/2) ★ p^4 +p^3 +2p+2=0 no useable exact solutions p≈−.975564±.528237i ⇒ x≈.33635∓.515329i ∧ y≈−.0559113±.680406i ∧ z=x ★ p≈.475564±1.18273i ⇒ x≈−.586346±.562464i ∧ y≈−.944089∓.426001i ∧ z=x ★ p^2 (q^2 +q+1)+q^4 +q^3 +q+1=0 p^2 =−(((q+1)^2 (q^2 −q+1))/(q^2 +q+1)) we can be sure that (q≠−1⇒p=0) ⇒ p^2 <0 ⇒ p=±(√((q^2 −q+1)/(q^2 +q+1)))(q+1)i ...I′ll continue later


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Question Number 176718 by Frix last updated on 25/Sep/22

$x^{3} + y^{3} = z^{2}$ $x^{3} + z^{3} = y^{2}$ $y^{3} + z^{3} = x^{2}$ $obviously x = y = z = 0 \lor \frac{1}{2}$ $trying to totally solve it$ $let y = p x \land z = q x$ $(p^{3} + 1) x^{3} = q^{2} x^{2}$ $(q^{3} + 1) x^{3} = p^{2} x^{2}$ $(p^{3} + q^{3}) x^{3} = x^{2}$ $\Rightarrow x = 0 \Rightarrow y = 0 \land z = 0 ★$ $x = \frac{q^{2}}{p^{3} + 1}$ $x = \frac{p^{2}}{q^{3} + 1}$ $x = \frac{1}{p^{3} + q^{3}}$ $\Rightarrow$ $\frac{p^{2}}{q^{3} + 1} = \frac{q^{2}}{p^{3} + 1}$ $\frac{p^{2}}{q^{3} + 1} = \frac{1}{p^{3} + q^{3}}$ $==========$ $p^{5} + p^{2} - q^{5} - q^{2} = 0$ $p^{5} + p^{2} q^{3} - q^{3} - 1 = 0$ $subtracting both$ $p^{2} (q^{3} - 1) + q^{5} - q^{3} + q^{2} - 1 = 0$ $(q - 1) (p^{2} (q^{2} + q + 1) + q^{4} + q^{3} + q + 1) = 0$ $\Rightarrow q = 1 \Rightarrow p^{5} + p^{2} - 2 = 0$ $(p - 1) (p^{4} + p^{3} + 2 p + 2) = 0$ $\Rightarrow p = 1 \Rightarrow x = y = z = \frac{1}{2} ★$ $p^{4} + p^{3} + 2 p + 2 = 0$ $no useable exact solutions$ $p \approx - . 975564 \pm . 528237 i$ $\Rightarrow x \approx . 33635 \mp . 515329 i$ $\land y \approx - . 0559113 \pm . 680406 i$ $\land z = x ★$ $p \approx . 475564 \pm 1 . 18273 i$ $\Rightarrow x \approx - . 586346 \pm . 562464 i$ $\land y \approx - . 944089 \mp . 426001 i$ $\land z = x ★$ $p^{2} (q^{2} + q + 1) + q^{4} + q^{3} + q + 1 = 0$ $p^{2} = - \frac{{(q + 1)}^{2} (q^{2} - q + 1)}{q^{2} + q + 1}$ $we can be sure that (q \neq - 1 \Rightarrow p = 0)$ $\Rightarrow p^{2} < 0 \Rightarrow p = \pm \sqrt{\frac{q^{2} - q + 1}{q^{2} + q + 1}} (q + 1) i$ $. . . I^{'} ll continue later$
Commented by Tawa11 last updated on 25/Sep/22

$Great sir$
	Answered by behi834171 last updated on 26/Sep/22

	$y^{3} - z^{3} = z^{2} - y^{2} \Rightarrow (y - z) (y^{2} + z^{2} + y z) = - (y - z) (y + z)$ $\Rightarrow x = y = z = 0 o r \frac{1}{2}$ $y^{2} + z^{2} + y z = - (y + z)$ $z^{2} + x^{2} + x z = - (x + z)$ $x^{2} + y^{2} + x y = - (x + y)$ $\Rightarrow y^{2} - x^{2} + z (y - x) = - (y - x) \Rightarrow$ $\Rightarrow (y - x) (y + x + z + 1) = 0$ $\Rightarrow y + x + z = - 1$ $. . .$
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