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Question Number 92717 by mr W last updated on 08/May/20

Commented by i jagooll last updated on 09/May/20

$(1) x^{3} + {9 x}^{2} y = 10$ $(2) {27 y}^{3} + {27 xy}^{2} = 54$ $(1) + (2)$ $x^{3} + {9 x}^{2} y + {27 xy}^{2} + {27 y}^{3} = 64$ ${(x + 3 y)}^{3} = 4^{3} \Rightarrow x + 3 y = 4$ $x = 4 - 3 y$ $\Rightarrow y^{3} + (4 - 3 y) y^{2} = 2$ $- {2 y}^{3} + {4 y}^{2} - 2 = 0$ $y^{3} - {2 y}^{2} + 1 = 0$ $(y - 1) (y^{2} - y - 1) = 0$ $y = 1 \Rightarrow x = 1$ $y = \frac{1 \pm \sqrt{5}}{2} \Rightarrow x = 4 - (\frac{3 \pm 3 \sqrt{5}}{2})$ $x = \frac{5 \mp 3 \sqrt{5}}{2}$
Commented by mr W last updated on 09/May/20

$r e s u l t n o t c o r r e c t s i r, i g o t :$ $x = 4 - 3 y$ $y^{3} + (4 - 3 y) y^{2} = 2$ $y^{3} - 2 y^{2} + 1 = 0$ $y^{3} - y^{2} - y^{2} + y - y + 1 = 0$ $(y - 1) y^{2} - (y - 1) y - (y - 1) = 0$ $(y - 1) (y^{2} - y - 1) = 0$ $y = 1 \Rightarrow x = 1$ $y = \frac{1 \pm \sqrt{5}}{2} \Rightarrow x = \frac{5 \mp 3 \sqrt{5}}{2}$
Commented by i jagooll last updated on 09/May/20

$dear Mr W . it correct ?$
Commented by i jagooll last updated on 09/May/20

$o yes sir . i typo$
Commented by john santu last updated on 09/May/20
greeatm..��
	Answered by Rasheed.Sindhi last updated on 08/May/20

	$x^{2} (x + 9 y) = 10$ $y^{2} (x + y) = 2$ $\frac{x^{2}}{y^{2}} \times (1 + \frac{8 y}{x + y}) = 5$ ${(\frac{x}{y})}^{2} (1 + \frac{1}{\frac{x + y}{8 y}}) = 5$ ${(\frac{x}{y})}^{2} (1 + \frac{1}{\frac{1}{8} (\frac{x}{y}) + \frac{1}{8}}) = 5$ $u^{2} (1 + \frac{1}{(1 / 8) (u + 1)}) = 5$ $u^{2} (\frac{(1 / 8) (u + 1) + 1}{(1 / 8) (u + 1)}) = 5$ $(1 / 8) u^{2} (u + 1) + u^{2} = (5 / 8) (u + 1)$ $u^{2} (u + 1) + 8 u^{2} = 5 (u + 1)$ $u^{3} + 9 u^{2} - 5 u - 5 = 0$ ${D o n}^{'} t k n o w t o s o l v e c u b i c e q u a t i o n .$ $. . . . . . . . .$ $. . . . .$
	Commented by Rio Michael last updated on 08/May/20

	$mind if i continue ?$ $u = 1 \Rightarrow {(1)}^{3} + 9 {(1)}^{2} - 5 (1) - (5) = 0$ $\Rightarrow if f (u) = u^{3} + 9 u^{2} - 5 u - 5$ $then u - 1 is a factor hence$ $f (u) = (u - 1) (u + 5 - \sqrt{20}) (u + 5 + \sqrt{20})$ $u = 1 or u = \sqrt{20} - 5 or u = - 5 - \sqrt{20}$
	Commented by mr W last updated on 08/May/20

	$t h a n k y o u b o t h!$
	Commented by Ar Brandon last updated on 09/May/20
	�� I love this spirit. Keep up Men.��
	Commented by Rasheed.Sindhi last updated on 13/May/20
	7ank5 5ir for encouraging!
	Answered by $@ty@m123 last updated on 10/May/20
	$((x^3 +9x^2 y)/(y^3 +xy^2 ))=5 ((((x/y))^3 +9((x/y))^2 )/(1+((x/y))))=5 t^3 +9t^2 −5t−5=0 {where t=(x/y) t^3 −t^2 +10t^2 −10t+5t−5=0 t^2 (t−1)+10t(t−1)+5(t−1)=0 (t−1)(t^2 +10t+5)=0 t=1, t=((−10±(√(100−20)))/2) t=1, t=−5±2(√5) Case 1. t=1 x=y=1. Case 2. t=−5+2(√5) ⇒x=(−5+2(√5))y ⇒y^3 +(−5+2(√5))y^3 =2 (−4+2(√5))y^3 =2 y^3 =(1/(−2+(√5))) y^3 =(1/((√5)−2))×(((√5)+2)/((√5)+2)) =(((√5)+2)/(5−4))=(√5)+2 y=((√5)+2)^(1/3) x=(−5+2(√5))((√5)+2)^(1/3) Case 3. t=−5−2(√5) ..... .....$
	$\frac{x^{3} + 9 x^{2} y}{y^{3} + {x y}^{2}} = 5$ $\frac{{(\frac{x}{y})}^{3} + 9 {(\frac{x}{y})}^{2}}{1 + (\frac{x}{y})} = 5$ $t^{3} + 9 t^{2} - 5 t - 5 = 0 {w h e r e t = \frac{x}{y}$ $t^{3} - t^{2} + 10 t^{2} - 10 t + 5 t - 5 = 0$ $t^{2} (t - 1) + 10 t (t - 1) + 5 (t - 1) = 0$ $(t - 1) (t^{2} + 10 t + 5) = 0$ $t = 1, t = \frac{- 10 \pm \sqrt{100 - 20}}{2}$ $t = 1, t = - 5 \pm 2 \sqrt{5}$ $C a s e 1 . t = 1$ $x = y = 1 .$ $C a s e 2 . t = - 5 + 2 \sqrt{5}$ $\Rightarrow x = (- 5 + 2 \sqrt{5}) y$ $\Rightarrow y^{3} + (- 5 + 2 \sqrt{5}) y^{3} = 2$ $(- 4 + 2 \sqrt{5}) y^{3} = 2$ $y^{3} = \frac{1}{- 2 + \sqrt{5}}$ $y^{3} = \frac{1}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2}$ $= \frac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2$ $y = {(\sqrt{5} + 2)}^{\frac{1}{3}}$ $x = (- 5 + 2 \sqrt{5}) {(\sqrt{5} + 2)}^{\frac{1}{3}}$ $C a s e 3 . t = - 5 - 2 \sqrt{5}$ $. . . . .$ $. . . . .$
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