((√2) +(√3) +1)((√2) −(√3) +1)((√2) +(√3)−1)((√2)−(√3)−1)=?


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Question Number 135300 by bobhans last updated on 12/Mar/21

$(\sqrt{2} + \sqrt{3} + 1) (\sqrt{2} - \sqrt{3} + 1) (\sqrt{2} + \sqrt{3} - 1) (\sqrt{2} - \sqrt{3} - 1) = ?$
	Answered by Olaf last updated on 12/Mar/21

	$(\sqrt{2} + \sqrt{3} + 1) (\sqrt{2} - \sqrt{3} - 1) = 2 - {(\sqrt{3} + 1)}^{2}$ $= - 2 - 2 \sqrt{3} = - 2 (1 + \sqrt{3})$ $(\sqrt{2} + \sqrt{3} - 1) (\sqrt{2} - \sqrt{3} + 1) = 2 - {(\sqrt{3} - 1)}^{2}$ $= - 2 + 2 \sqrt{3} = - 2 (1 - \sqrt{3})$ $Total product :$ $4 (1 - 3) = - 8$
	Answered by john_santu last updated on 12/Mar/21
	$via Trigonometry let ((√2)+(√3)+1)((√2)−(√3)+1)((√2)+(√3)−1)((√2)−(√3)−1)=p then (((√2)/2)+((√3)/2)+(1/2))(((√2)/2)−((√3)/2)+(1/2))(((√2)/2)+((√3)/2)−(1/2))(((√2)/2)−((√3)/2)−(1/2))=(p/(16)) consider : ((√3)/2)+(1/2)= cos (π/6)+cos (π/3) = 2cos (π/4)cos (π/(12))=(√2) cos (π/(12)) for the first factor ((√2)/2)(1+2cos (π/(12))) for the last factor ((√2)/2)(1−2cos (π/(12))) so the product of these two factors will be (1/2)(1−4cos^2 (π/(12))) likewise for difference of the relevant summands we shall have (1/2)(1−4sin^2 (π/(12))) →now { ((4sin^2 (π/(12)) = 2−(√3))),((4cos^2 (π/(12)) = 2+(√3))) :} ∴ (1/2)(1−4cos^2 (π/(12)))(1−4sin^2 (π/(12)))=(p/(16)) ∴ (1/4)(−1)((√3)−1)((√3)+1)=(p/(16)) ∴ p = 16×(−(1/2))=−8 [ ((√2)+(√3)+1)_(((√2)/2)(1+2cos (π/(12)))) ((√2)−(√3)+1)_(((√2)/2)(1−2cos (π/(12)))) ((√2)+(√3)−1)_(((√2)/2)(1+2sin (π/(12)))) ((√2)−(√3)−1)_(((√2)/2)(1−2sin (π/(12)))) ]$
	$v i a T r i g o n o m e t r y$ $l e t (\sqrt{2} + \sqrt{3} + 1) (\sqrt{2} - \sqrt{3} + 1) (\sqrt{2} + \sqrt{3} - 1) (\sqrt{2} - \sqrt{3} - 1) = p$ $t h e n (\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{1}{2}) (\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} + \frac{1}{2}) (\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} - \frac{1}{2}) (\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} - \frac{1}{2}) = \frac{p}{16}$ $c o n s i d e r : \frac{\sqrt{3}}{2} + \frac{1}{2} = \cos \frac{π}{6} + \cos \frac{π}{3}$ $= 2 \cos \frac{π}{4} \cos \frac{π}{12} = \sqrt{2} \cos \frac{π}{12}$ $f o r t h e f i r s t f a c t o r \frac{\sqrt{2}}{2} (1 + 2 \cos \frac{π}{12})$ $f o r t h e l a s t f a c t o r \frac{\sqrt{2}}{2} (1 - 2 \cos \frac{π}{12})$ $s o t h e p r o d u c t o f t h e s e t w o f a c t o r s$ $w i l l b e \frac{1}{2} (1 - 4 \cos^{2} \frac{π}{12})$ $l i k e w i s e f o r d i f f e r e n c e o f t h e$ $r e l e v a n t s u m m a n d s w e s h a l l h a v e$ $\frac{1}{2} (1 - 4 \sin^{2} \frac{π}{12})$ $\to n o w {\begin{cases} 4 \sin^{2} \frac{π}{12} = 2 - \sqrt{3} \\ 4 \cos^{2} \frac{π}{12} = 2 + \sqrt{3} \end{cases}$ $∴ \frac{1}{2} (1 - 4 \cos^{2} \frac{π}{12}) (1 - 4 \sin^{2} \frac{π}{12}) = \frac{p}{16}$ $∴ \frac{1}{4} (- 1) (\sqrt{3} - 1) (\sqrt{3} + 1) = \frac{p}{16}$ $∴ p = 16 \times (- \frac{1}{2}) = - 8$ $[\underset{\frac{\sqrt{2}}{2} (1 + 2 \cos \frac{π}{12})}{\underset{⏟}{(\sqrt{2} + \sqrt{3} + 1)}} \underset{\frac{\sqrt{2}}{2} (1 - 2 \cos \frac{π}{12})}{\underset{⏟}{(\sqrt{2} - \sqrt{3} + 1)}} \underset{\frac{\sqrt{2}}{2} (1 + 2 \sin \frac{π}{12})}{\underset{⏟}{(\sqrt{2} + \sqrt{3} - 1)}} \underset{\frac{\sqrt{2}}{2} (1 - 2 \sin \frac{π}{12})}{\underset{⏟}{(\sqrt{2} - \sqrt{3} - 1)}}]$
	Answered by som(math1967) last updated on 12/Mar/21
	$let (√2)=a,(√3)=b,1=c (a+b+c)(a−b+c)(a+b−c)(a−b−c) ={(a+c)^2 −b^2 }{(a−c)^2 −b^2 } =(a^2 −b^2 +c^2 +2ac)(a^2 +c^2 −b^2 −2ac) =(a^2 −b^2 +c^2 )^2 −4a^2 c^2 =a^4 +b^4 +c^4 −2a^2 b^2 −2b^2 c^2 −2a^2 c^2 =4+9+1−2.2.3−2.3.1−2.2.1 =14−22=−8$
	$l e t \sqrt{2} = a, \sqrt{3} = b, 1 = c$ $(a + b + c) (a - b + c) (a + b - c) (a - b - c)$ $= {{(a + c)}^{2} - b^{2}} {{(a - c)}^{2} - b^{2}}$ $= (a^{2} - b^{2} + c^{2} + 2 a c) (a^{2} + c^{2} - b^{2} - 2 a c)$ $= {(a^{2} - b^{2} + c^{2})}^{2} - 4 a^{2} c^{2}$ $= a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 b^{2} c^{2} - 2 a^{2} c^{2}$ $= 4 + 9 + 1 - 2.2.3 - 2.3.1 - 2.2.1$ $= 14 - 22 = - 8$
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