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Question Number 183864 by Michaelfaraday last updated on 31/Dec/22
Commented by MJS_new last updated on 31/Dec/22
b)
$$\left.{b}\right) \\ $$
Commented by Michaelfaraday last updated on 31/Dec/22
sir,show workings
$${sir},{show}\:{workings} \\ $$
Answered by mr W last updated on 31/Dec/22
((√(11))+(√5))^2 =16+2(√(55)) ((√(11))+(√5))^4 =4(119+16(√(55))) ((√(11))+(√5))^8 =16(119^2 +16^2 ×55+2×119×16(√(55))) similarly ((√(11))−(√5))^8 =16(119^2 +16^2 ×55−2×119×16(√(55))) ((√(11))+(√5))^8 +((√(11))−(√5))^8 =2×16(119^2 +16^2 ×55) =903712 ✓
$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{16}+\mathrm{2}\sqrt{\mathrm{55}} \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{4}} =\mathrm{4}\left(\mathrm{119}+\mathrm{16}\sqrt{\mathrm{55}}\right) \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\mathrm{16}\left(\mathrm{119}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} ×\mathrm{55}+\mathrm{2}×\mathrm{119}×\mathrm{16}\sqrt{\mathrm{55}}\right) \\ $$$${similarly} \\ $$$$\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\mathrm{16}\left(\mathrm{119}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} ×\mathrm{55}−\mathrm{2}×\mathrm{119}×\mathrm{16}\sqrt{\mathrm{55}}\right) \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\mathrm{2}×\mathrm{16}\left(\mathrm{119}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} ×\mathrm{55}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{903712}\:\checkmark \\ $$
Commented by Michaelfaraday last updated on 31/Dec/22
thanks sir
$${thanks}\:{sir} \\ $$
Answered by Acem last updated on 31/Dec/22
k= ((√(11))+(√5))^8 + ((√(11))−(√5))^8 = 11^4 [(1+(√(5/(11))))^8 +(1−(√(5/(11))))^8 ] δ= (√(5/(11))) k= 11^4 [(1+δ)^8 +(1−δ)^8 ] = 11^4 [2(1+ ((8),(2) ) δ^( 2) + ((8),(4) ) δ^( 4) + ((8),(6) ) δ^( 6) +δ^( 8 ) )] = 11^4 [ 2 (1+ (5/(11)) (28 + 70 (5/(11)) + 28 (5^2 /(11^2 )) + (5^3 /(11^3 )) ) ] = 903 712
$${k}=\:\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} +\:\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\:\mathrm{11}^{\mathrm{4}} \left[\left(\mathrm{1}+\sqrt{\frac{\mathrm{5}}{\mathrm{11}}}\right)^{\mathrm{8}} +\left(\mathrm{1}−\sqrt{\frac{\mathrm{5}}{\mathrm{11}}}\right)^{\mathrm{8}} \right] \\ $$$$\:\delta=\:\sqrt{\frac{\mathrm{5}}{\mathrm{11}}} \\ $$$$\:{k}=\:\mathrm{11}^{\mathrm{4}} \left[\left(\mathrm{1}+\delta\right)^{\mathrm{8}} +\left(\mathrm{1}−\delta\right)^{\mathrm{8}} \right] \\ $$$$\:\:\:=\:\mathrm{11}^{\mathrm{4}} \left[\mathrm{2}\left(\mathrm{1}+\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:\delta^{\:\mathrm{2}} +\begin{pmatrix}{\mathrm{8}}\\{\mathrm{4}}\end{pmatrix}\:\delta^{\:\mathrm{4}} +\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{6}}\end{pmatrix}\:\delta^{\:\mathrm{6}} +\delta^{\:\mathrm{8}\:} \right)\right] \\ $$$$\:\:\:=\:\mathrm{11}^{\mathrm{4}} \:\left[\:\mathrm{2}\:\left(\mathrm{1}+\:\frac{\mathrm{5}}{\mathrm{11}}\:\left(\mathrm{28}\:+\:\mathrm{70}\:\frac{\mathrm{5}}{\mathrm{11}}\:+\:\mathrm{28}\:\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{11}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{11}^{\mathrm{3}} }\:\right)\:\right]\right. \\ $$$$\:\:\:=\:\mathrm{903}\:\mathrm{712} \\ $$$$ \\ $$
Commented by Michaelfaraday last updated on 31/Dec/22
thanks sir
$${thanks}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 31/Dec/22
a=(√(11)) +(√5) , b=(√(11)) −(√5) a+b=2(√(11)) , ab=11−5=6 a^2 +b^2 =(a+b)^2 −2ab =(2(√(11)) )^2 −2(6)=44−12=10 a^4 +b^4 =(a^2 +b^2 )^2 −2a^2 b^2 =32^2 −2(6)^2 =952 a^8 +b^8 =(a^4 +b^4 )^2 −2a^4 b^4 =952^2 −2(6)^4 =903712
$${a}=\sqrt{\mathrm{11}}\:+\sqrt{\mathrm{5}}\:\:,\:\:{b}=\sqrt{\mathrm{11}}\:−\sqrt{\mathrm{5}}\: \\ $$$$\:\:\:\:\:\:\:\:{a}+{b}=\mathrm{2}\sqrt{\mathrm{11}}\:,\:{ab}=\mathrm{11}−\mathrm{5}=\mathrm{6} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}\sqrt{\mathrm{11}}\:\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{6}\right)=\mathrm{44}−\mathrm{12}=\mathrm{10} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{32}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{6}\right)^{\mathrm{2}} =\mathrm{952} \\ $$$${a}^{\mathrm{8}} +{b}^{\mathrm{8}} =\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{952}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{6}\right)^{\mathrm{4}} =\mathrm{903712} \\ $$
Commented by Acem last updated on 31/Dec/22
Also cool
$${Also}\:{cool} \\ $$
Commented by manxsol last updated on 31/Dec/22
very very good
$${very}\:{very}\:{good} \\ $$
Commented by Rasheed.Sindhi last updated on 31/Dec/22
ThanX Acem & manxsol sirs!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\mathbb{X}\:{Acem}\:\&\:{manxsol}\:{sirs}! \\ $$
Commented by Michaelfaraday last updated on 31/Dec/22
thanks sir
$${thanks}\:{sir} \\ $$
Answered by manxsol last updated on 31/Dec/22
((√(11))+(√5))^2 +((√(11))−(√5))^2 =2(16)=32 (a+b)^2 +(a−b)^2 =2(a^2 +b^2 ) ((√(11))+(√5))^4 +((√(11))−(√5))^4 +2(11−5)^2 =32^2 2((√(11))+(√5))^2 ((√(11))+(√5))^2 =2[(√(11))^2 +(√(5 ))^2 ]^2 =2(11−5)^2 ((√(11))+(√5))^4 +((√(11))−(√5))^4 =32^2 −2×6^2 =952 ((√(11))+(√5))^8 +((√(11))−(√5))^8 +2(11−5)^4 =952^2 ((√(11))+(√5))^8 +((√(11))−(√5))^8 =952^2 −2(6)^4 ((√(11))+(√5))^8 +((√(11))−(√5))^8 = 952^2 −2×6^4 903712.0
$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{16}\right)=\mathrm{32} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{4}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{4}} +\mathrm{2}\left(\mathrm{11}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{32}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{2}\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{2}\left[\sqrt{\mathrm{11}}\:^{\mathrm{2}} +\sqrt{\mathrm{5}\:}\:^{\mathrm{2}} \right]^{\mathrm{2}} =\mathrm{2}\left(\mathrm{11}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{4}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{4}} =\mathrm{32}^{\mathrm{2}} −\mathrm{2}×\mathrm{6}^{\mathrm{2}} =\mathrm{952} \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{8}} +\mathrm{2}\left(\mathrm{11}−\mathrm{5}\right)^{\mathrm{4}} =\mathrm{952}^{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\mathrm{952}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{6}\right)^{\mathrm{4}} \\ $$$$\left(\sqrt{\mathrm{11}}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} +\left(\sqrt{\mathrm{11}}−\sqrt{\mathrm{5}}\right)^{\mathrm{8}} = \\ $$$$\mathrm{952}^{\mathrm{2}} −\mathrm{2}×\mathrm{6}^{\mathrm{4}} \\ $$$$\mathrm{903712}.\mathrm{0} \\ $$$$ \\ $$
Commented by Michaelfaraday last updated on 31/Dec/22
thanks sir
$${thanks}\:{sir} \\ $$

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