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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

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

Question Number 176684 Answers: 1 Comments: 2

Question Number 176676 Answers: 3 Comments: 0

If x^3 +(1/x^3 )=1, prove that x^5 +(1/x^5 )=−(x^4 +(1/x^4 ))

$If x^{3} + \frac{1}{x^{3}} = 1,$ $prove that$ $x^{5} + \frac{1}{x^{5}} = - (x^{4} + \frac{1}{x^{4}})$

Question Number 176636 Answers: 2 Comments: 2

{ ((p^3 +q^3 =r^2 )),((p^3 +r^3 =q^2 )),((q^3 +r^3 =p^2 )) :} ⇒20pqr =?

${\begin{cases} p^{3} + q^{3} = r^{2} \\ p^{3} + r^{3} = q^{2} \\ q^{3} + r^{3} = p^{2} \end{cases}$ $\Rightarrow 20 p q r = ?$

Question Number 176592 Answers: 2 Comments: 0

Solve the equation: 2^x + 3^x − 4^x + 6^x − 9^x = 1

$Solve the equation :$ $2^{x} + 3^{x} - 4^{x} + 6^{x} - 9^{x} = 1$

Question Number 176598 Answers: 1 Comments: 0

x^3 +(1/x^3 )=1 (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=? Q#176387 reposted for a new answer.

$x^{3} + \frac{1}{x^{3}} = 1$ $\frac{{(x^{5} + \frac{1}{x^{5}})}^{3} - 1}{x^{5} + \frac{1}{x^{5}}} = ?$ $You can't use 'macro parameter character #' in math mode$

Question Number 176580 Answers: 1 Comments: 0

4^(x^2 −2x+2) −2^(x^2 −2x+3) +2=2^(x^2 −2x+2) x=?

$4^{x^{2} - 2 x + 2} - 2^{x^{2} - 2 x + 3} + 2 = 2^{x^{2} - 2 x + 2}$ $x = ?$

Question Number 176555 Answers: 1 Comments: 2

ABCD−convex quadrilateral M∈Int(ABCD) , F−area , s−semiperimetr a , b , c−sides. Prove that: ((MA^4 )/b) + ((MB^4 )/c^4 ) + ((MC^4 )/d) + ((MD^4 )/a) ≥ ((2F^2 )/s)

$ABCD - convex quadrilateral$ $M \in Int (ABCD), F - area, s - semiperimetr$ $a, b, c - sides . Prove that :$ $\frac{{MA}^{4}}{b} + \frac{{MB}^{4}}{c^{4}} + \frac{{MC}^{4}}{d} + \frac{{MD}^{4}}{a} ⩾ \frac{{2 F}^{2}}{s}$

Question Number 176554 Answers: 2 Comments: 0

x + (1/x) =ϕ (Golden ratio) then x^( 2000) +(( 1)/x^( 2000) ) = ?

$x + \frac{1}{x} = φ (G o l d e n r a t i o)$ $t h e n$ $x^{2000} + \frac{1}{x^{2000}} = ?$

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