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Question Number 161946 by mathlove last updated on 24/Dec/21

	Answered by Rasheed.Sindhi last updated on 29/Dec/21

	$x + y (x + 1) = 3 \Rightarrow y = \frac{3 - x}{x + 1}$ $\Rightarrow y + 1 = \frac{3 - x + x + 1}{x + 1} = \frac{4}{x + 1}$ $y + z (y + 1) = 5 \Rightarrow \frac{3 - x}{x + 1} + z (\frac{4}{x + 1}) = 5$ $\Rightarrow 3 - x + 4 z = 5 x + 5 \Rightarrow x = \frac{2 z - 1}{3}$ $z + x (z + 1) = 7 \Rightarrow z + (\frac{2 z - 1}{3}) (z + 1) = 7$ $\Rightarrow 3 z + 2 z^{2} + z - 1 = 21$ $z^{2} + 2 z - 11 = 0$ $z = \frac{- 2 \pm \sqrt{4 + 44}}{2} = \frac{- 2 \pm 4 \sqrt{3}}{2} = - 1 \pm 2 \sqrt{3}$ $x = \frac{2 z - 1}{3} = \frac{2 (- 1 \pm 2 \sqrt{3}) - 1}{3} = \frac{- 3 \pm 4 \sqrt{3}}{3}$ $y = \frac{3 - x}{x + 1} = \frac{3 - (\frac{- 3 \pm 4 \sqrt{3}}{3})}{(\frac{- 3 \pm 4 \sqrt{3}}{3}) + 1}$ $= \frac{9 + 3 \mp 4 \sqrt{3}}{\pm 4 \sqrt{3}} \times \frac{\pm 4 \sqrt{3}}{\pm 4 \sqrt{3}} = \frac{\pm 48 \sqrt{3} - 48}{48}$ $y = - 1 \pm \sqrt{3}$ $x + y + z = (\frac{- 3 \pm 4 \sqrt{3}}{3}) + (- 1 \pm \sqrt{3}) + (- 1 \pm 2 \sqrt{3})$ $= \frac{- 3 \pm 4 \sqrt{3} - 3 \pm 3 \sqrt{3} - 3 \pm 6 \sqrt{3}}{3}$ $= \frac{- 9 \pm 13 \sqrt{3}}{3}$ ${(x + y + z)}^{2} = {(\frac{- 9 \pm 13 \sqrt{3}}{3})}^{2}$ $= \frac{81 \mp 234 \sqrt{3} + 507}{9}$ $= \frac{588 \mp 234 \sqrt{3}}{9} = \frac{196 \mp 78 \sqrt{3}}{3}$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$\underset{\underset{L \underset{T}{\overset{L}{O} T}}{A}}{T^{H^{A} N} X}$ $S \underset{\underset{∙}{I}}{\overset{\underset{}{∙}}{I}} R$
	Answered by mr W last updated on 29/Dec/21
	$let u=x+1 v=y+1 w=z+1 eqn. (i) becomes u−1+v−1+(u−1)(v−1)=3 uv=4 ...(I) similarly vw=6 ...(II) wu=8 ...(III) (I)×(II)×(III): (uvw)^2 =4×6×8=192 uvw=±(√(192))=±8(√3) ...(IV) (IV)/(I): w=±((8(√3))/4)=±2(√3)=z+1 ⇒z=−1±2(√3) similarly u=±((8(√3))/6)=±((4(√3))/3)=x+1 ⇒x=−1±((4(√3))/3) v=±((8(√3))/8)=±(√3)=y+1 ⇒y=−1±(√3) (x+y+z)^2 =(u+v+w−3)^2 =(±((4(√3))/3)±(√3)±2(√3)−3)^2 =(±((13(√3))/3)−3)^2 = { (((196−78(√3))/3)),(((196+78(√3))/3)) :}$
	$l e t$ $u = x + 1$ $v = y + 1$ $w = z + 1$ $e q n . (i) b e c o m e s$ $u - 1 + v - 1 + (u - 1) (v - 1) = 3$ $u v = 4 . . . (I)$ $s i m i l a r l y$ $v w = 6 . . . (I I)$ $w u = 8 . . . (I I I)$ $(I) \times (I I) \times (I I I) :$ ${(u v w)}^{2} = 4 \times 6 \times 8 = 192$ $u v w = \pm \sqrt{192} = \pm 8 \sqrt{3} . . . (I V)$ $(I V) / (I) :$ $w = \pm \frac{8 \sqrt{3}}{4} = \pm 2 \sqrt{3} = z + 1 \Rightarrow z = - 1 \pm 2 \sqrt{3}$ $s i m i l a r l y$ $u = \pm \frac{8 \sqrt{3}}{6} = \pm \frac{4 \sqrt{3}}{3} = x + 1 \Rightarrow x = - 1 \pm \frac{4 \sqrt{3}}{3}$ $v = \pm \frac{8 \sqrt{3}}{8} = \pm \sqrt{3} = y + 1 \Rightarrow y = - 1 \pm \sqrt{3}$ ${(x + y + z)}^{2} = {(u + v + w - 3)}^{2} = {(\pm \frac{4 \sqrt{3}}{3} \pm \sqrt{3} \pm 2 \sqrt{3} - 3)}^{2}$ $= {(\pm \frac{13 \sqrt{3}}{3} - 3)}^{2} = {\begin{cases} \frac{196 - 78 \sqrt{3}}{3} \\ \frac{196 + 78 \sqrt{3}}{3} \end{cases}$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$Sir I {didn}^{'} t understand last line .$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$SMART METHOD SiR!$
	Commented by mr W last updated on 29/Dec/21

	$i t o n l y s h o w s t h a t t h e r e a r e t w o$ $v a l u e s .$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$Again Thanx!$
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