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Question Number 135300 by bobhans last updated on 12/Mar/21
((√2) +(√3) +1)((√2) −(√3) +1)((√2) +(√3)−1)((√2)−(√3)−1)=?
$$\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)=? \\ $$
Answered by Olaf last updated on 12/Mar/21
((√2)+(√3)+1)((√2)−(√3)−1) = 2−((√3)+1)^2 = −2−2(√3) = −2(1+(√3)) ((√2)+(√3)−1)((√2)−(√3)+1) = 2−((√3)−1)^2 = −2+2(√3) = −2(1−(√3)) Total product : 4(1−3) = −8
$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\:=\:−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\:=\:−\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$$$\mathrm{Total}\:\mathrm{product}\:: \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{3}\right)\:=\:−\mathrm{8} \\ $$
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)))) ]
$${via}\:{Trigonometry} \\ $$$${let}\:\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)={p} \\ $$$${then}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{{p}}{\mathrm{16}} \\ $$$$ \\ $$$${consider}\::\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$$$=\:\mathrm{2cos}\:\frac{\pi}{\mathrm{4}}\mathrm{cos}\:\frac{\pi}{\mathrm{12}}=\sqrt{\mathrm{2}}\:\mathrm{cos}\:\frac{\pi}{\mathrm{12}} \\ $$$${for}\:{the}\:{first}\:{factor}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2cos}\:\frac{\pi}{\mathrm{12}}\right) \\ $$$${for}\:{the}\:{last}\:{factor}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2cos}\:\frac{\pi}{\mathrm{12}}\right) \\ $$$${so}\:{the}\:{product}\:{of}\:{these}\:{two}\:{factors} \\ $$$${will}\:{be}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{12}}\right) \\ $$$${likewise}\:{for}\:{difference}\:{of}\:{the}\: \\ $$$${relevant}\:{summands}\:{we}\:{shall}\:{have} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{4sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{12}}\right)\: \\ $$$$\rightarrow{now}\:\begin{cases}{\mathrm{4sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{12}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}}}\\{\mathrm{4cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{12}}\:=\:\mathrm{2}+\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{12}}\right)\left(\mathrm{1}−\mathrm{4sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{12}}\right)=\frac{{p}}{\mathrm{16}} \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)=\frac{{p}}{\mathrm{16}} \\ $$$$\therefore\:{p}\:=\:\mathrm{16}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{8} \\ $$$$\left[\:\underset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2cos}\:\frac{\pi}{\mathrm{12}}\right)} {\underbrace{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)}}\:\underset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2cos}\:\frac{\pi}{\mathrm{12}}\right)} {\underbrace{\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)}}\:\underset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2sin}\:\frac{\pi}{\mathrm{12}}\right)} {\underbrace{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)}}\:\underset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2sin}\:\frac{\pi}{\mathrm{12}}\right)} {\underbrace{\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)}}\:\right] \\ $$$$ \\ $$
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
$${let}\:\sqrt{\mathrm{2}}={a},\sqrt{\mathrm{3}}={b},\mathrm{1}={c} \\ $$$$\left({a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}−{c}\right) \\ $$$$=\left\{\left({a}+{c}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} \right\}\left\{\left({a}−{c}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} \right\} \\ $$$$=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\right)\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{ac}\right) \\ $$$$=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$=\mathrm{4}+\mathrm{9}+\mathrm{1}−\mathrm{2}.\mathrm{2}.\mathrm{3}−\mathrm{2}.\mathrm{3}.\mathrm{1}−\mathrm{2}.\mathrm{2}.\mathrm{1} \\ $$$$=\mathrm{14}−\mathrm{22}=−\mathrm{8} \\ $$

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