Is there any solution(s)?? { ((36x^2 y−27y^3 =8)),((4x^3 −27xy^2 =4)) :} please....


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Question Number 123114 by nimnim last updated on 23/Nov/20
$Is there any solution(s)?? { ((36x^2 y−27y^3 =8)),((4x^3 −27xy^2 =4)) :} please....$
$I s t h e r e a n y s o l u t i o n (s) ? ?$ ${\begin{cases} 36 x^{2} y - 27 y^{3} = 8 \\ 4 x^{3} - 27 {x y}^{2} = 4 \end{cases}$ $p l e a s e . . . .$
Commented by benjo_mathlover last updated on 23/Nov/20
${ ((36x^2 y−27y^3 =8)),((8x^3 −54xy^2 =8)) :} ⇒36x^2 y−8x^3 +54xy^2 −27y^3 =0 ⇒x^2 (36y−8x)+27y^2 (2x−y)=0 ⇒x^2 (36y−8x)=27y^2 (y−2x) ⇒((x/y))^2 (36−((8x)/y))=27(1−2((x/y))) let (x/y) = r ⇒r^2 (36−8r)=27(1−2r) ⇒36r^2 −8r^3 = 27−54r ⇒8r^3 −36r^2 −54r+27=0 ⇒(2r+3)(4r^2 −24r+9)=0 r_1 = −(3/2)⇒(x/y)=−(3/2) or 2x=−3y r_(2,3) = −3±((3(√3))/2)$
${\begin{cases} 36 x^{2} y - 27 y^{3} = 8 \\ 8 x^{3} - 54 {x y}^{2} = 8 \end{cases}$ $\Rightarrow 36 x^{2} y - 8 x^{3} + 54 {x y}^{2} - 27 y^{3} = 0$ $\Rightarrow x^{2} (36 y - 8 x) + 27 y^{2} (2 x - y) = 0$ $\Rightarrow x^{2} (36 y - 8 x) = 27 y^{2} (y - 2 x)$ $\Rightarrow {(\frac{x}{y})}^{2} (36 - \frac{8 x}{y}) = 27 (1 - 2 (\frac{x}{y}))$ $l e t \frac{x}{y} = r$ $\Rightarrow r^{2} (36 - 8 r) = 27 (1 - 2 r)$ $\Rightarrow 36 r^{2} - 8 r^{3} = 27 - 54 r$ $\Rightarrow 8 r^{3} - 36 r^{2} - 54 r + 27 = 0$ $\Rightarrow (2 r + 3) (4 r^{2} - 24 r + 9) = 0$ $r_{1} = - \frac{3}{2} \Rightarrow \frac{x}{y} = - \frac{3}{2} o r 2 x = - 3 y$ $r_{2, 3} = - 3 \pm \frac{3 \sqrt{3}}{2}$
Commented by nimnim last updated on 23/Nov/20

$T h a n k y o u S i r .$
	Answered by mindispower last updated on 23/Nov/20
	$⇔multiplie (1).i { ((36ix^2 y−27iy^3 =8i..(1))),((8x^3 −54xy^2 =8....(2))) :} and Eq 2 ∗2 (1)+(2)⇒−27iy^3 +8x^3 −54xy^2 +36ix^2 y=16 ⇔(3iy)^3 +(2x)^3 +3(3iy)^2 2+3(3iy)(2x)^2 =8(i+1) ⇔(3iy+2x)^3 =8(√2)e^(((iπ)/4)+i2kπ) ....x,y∈R this is suffisant to solve it if x,y∈C (1)∗−i+...(2)∗2 ⇒(−3iy)^3 +3(−3iy)(2x)^2 +(2x)^3 +3(−3iy)^2 .2x=−8i+8 ⇒(2x−3iy)^3 =8(1−i)=8(√2)e^(−((iπ)/4)+2k_1 π) ,k_1 ∈{0,1,2} ⇔ { ((2x−3iy=8(√2)e^(−i(π/(12))+2((ikπ)/3)) )),((2x+3iy=8(√2)e^(i(π/(12))+2((k_1 π)/3)) )) :} solve and theck evry solution because We used ⇒$
	$\Leftrightarrow m u l t i p l i e (1) . i {\begin{cases} 36 {i x}^{2} y - 27 {i y}^{3} = 8 i . . (1) \\ 8 x^{3} - 54 {x y}^{2} = 8 . . . . (2) \end{cases}$ $a n d E q 2 * 2$ $(1) + (2) \Rightarrow - 27 {i y}^{3} + 8 x^{3} - 54 {x y}^{2} + 36 {i x}^{2} y = 16$ $\Leftrightarrow {(3 i y)}^{3} + {(2 x)}^{3} + 3 {(3 i y)}^{2} 2 + 3 (3 i y) {(2 x)}^{2} = 8 (i + 1)$ $\Leftrightarrow {(3 i y + 2 x)}^{3} = 8 \sqrt{2} e^{\frac{i π}{4} + i 2 k π} . . . . x, y \in R$ $t h i s i s s u f f i s a n t t o s o l v e i t$ $i f x, y \in C$ $(1) * - i + . . . (2) * 2$ $\Rightarrow {(- 3 i y)}^{3} + 3 (- 3 i y) {(2 x)}^{2} + {(2 x)}^{3} + 3 {(- 3 i y)}^{2} . 2 x = - 8 i + 8$ $\Rightarrow {(2 x - 3 i y)}^{3} = 8 (1 - i) = 8 \sqrt{2} e^{- \frac{i π}{4} + 2 k_{1} π}, k_{1} \in {0, 1, 2}$ $\Leftrightarrow {\begin{cases} 2 x - 3 i y = 8 \sqrt{2} e^{- i \frac{π}{12} + 2 \frac{i k π}{3}} \\ 2 x + 3 i y = 8 \sqrt{2} e^{i \frac{π}{12} + 2 \frac{k_{1} π}{3}} \end{cases}$ $s o l v e a n d t h e c k e v r y s o l u t i o n b e c a u s e W e$ $u s e d \Rightarrow$
	Commented by nimnim last updated on 23/Nov/20

	$T h a n k y o u S i r .$
	Answered by MJS_new last updated on 23/Nov/20

	$the easy path$ $36 x^{2} y - 27 y^{3} = 8$ $4 x^{3} - 27 {x y}^{2} = 4$ $let y = p x$ $36 {p x}^{3} - 27 p^{3} x^{3} = 8 \Leftrightarrow x^{3} = \frac{8}{36 p - 27 p^{3}}$ $4 x^{3} - 27 p^{2} x^{3} = 4 \Leftrightarrow x^{3} = \frac{4}{4 - 27 p^{2}}$ $\Rightarrow$ $36 p - 27 p^{3} = 8 - 54 p^{2}$ $p^{3} - 2 p^{2} - \frac{4}{3} p + \frac{8}{27} = 0$ $p_{1} = - \frac{2}{3}$ $p_{2} = \frac{4 - 2 \sqrt{3}}{3}$ $p_{3} = \frac{4 + 2 \sqrt{3}}{3}$ $\Rightarrow$ $x_{1} = - \frac{\sqrt[3]{4}}{2} \land y_{1} = \frac{\sqrt[3]{4}}{3}$ $x_{2} = \frac{\sqrt[3]{4}}{4} (1 + \sqrt{3}) \land y_{2} = \frac{\sqrt[3]{4}}{6} (- 1 + \sqrt{3})$ $x_{3} = - \frac{\sqrt[3]{4}}{4} (- 1 + \sqrt{3}) \land y_{3} = - \frac{\sqrt[3]{4}}{6} (1 + \sqrt{3})$
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