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Question Number 27002 by math solver last updated on 01/Jan/18

Commented by moxhix last updated on 01/Jan/18

x≠0 (∵0^2 +0+1≠0) ((x^2 +x+1)/x)=(0/x) x+(1/x)=−1 ...

$${x}\neq\mathrm{0}\:\left(\because\mathrm{0}^{\mathrm{2}} +\mathrm{0}+\mathrm{1}\neq\mathrm{0}\right) \\ $$$$\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}}=\frac{\mathrm{0}}{{x}} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=−\mathrm{1} \\ $$$$... \\ $$

Commented by math solver last updated on 01/Jan/18

what is the final ans.?

$${what}\:{is}\:{the}\:{final}\:{ans}.? \\ $$

Commented by abdo imad last updated on 01/Jan/18

the roots of x^2 +x+1=0 are j= e^(i((2π)/3)) and j^− = e^(−i((2π)/3)) let put S= (x+(1/x))^3 +(x^2 +(1/x^2 ))^3 +....( x^(100) +(1/x^(100) ))^3 S= Σ_(k=1) ^(100) ( j^k +j^k^− )^3 =Σ_(k=1) ^(k=100) (2Re(j^k ))^3 =8 Σ_(k=1) ^(k=100) (cos(2((kπ)/3)))^3 cos^3 x=(1/2)(1+cos(2x)cosx=(1/2)cosx +(1/4)(cos(3x) +cosx) =(3/4) cosx +(1/4) cos(3x) S=6Σ_(k=1) ^(100) cos(2((kπ)/3)) +2Σ_(k=1) ^(100) cos(2kπ) S= 200 +6 Re(Σ_(k=1) ^(100) e^(i((2kπ)/3)) ) but Σ_(k=1) ^(100) e^(i((2π)/3)k) = Σ_(k=0) ^(100) e^(i((2π)/3)k) −1 = ((1−(e^(i((2π)/3)) )^(101) )/(1−e^i^((2π)/3) )) −1 =((1− e^(i^((2π)/3) (99+2)) )/(1−e^(i((2π)/3)) )) −1= ((1−j^2 )/(1−j)) −1= 1+j −1=j and Re(Σ(..))=−(1/2) ⇒ S=200 −3=197.

$${the}\:{roots}\:{of}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:{are}\:{j}=\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{and}\:{j}^{−} =\:{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${let}\:{put}\:{S}=\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{3}} +....\left(\:{x}^{\mathrm{100}} +\frac{\mathrm{1}}{{x}^{\mathrm{100}} }\right)^{\mathrm{3}} \\ $$$${S}=\:\sum_{{k}=\mathrm{1}} ^{\mathrm{100}} \left(\:{j}^{{k}} +{j}^{{k}^{−} } \:\right)^{\mathrm{3}} \:=\sum_{{k}=\mathrm{1}} ^{{k}=\mathrm{100}} \left(\mathrm{2}{Re}\left({j}^{{k}} \:\right)\right)^{\mathrm{3}} \\ $$$$=\mathrm{8}\:\sum_{{k}=\mathrm{1}} ^{{k}=\mathrm{100}} \left({cos}\left(\mathrm{2}\frac{{k}\pi}{\mathrm{3}}\right)\right)^{\mathrm{3}} \\ $$$${cos}^{\mathrm{3}} {x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right){cosx}=\frac{\mathrm{1}}{\mathrm{2}}{cosx}\:+\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\left(\mathrm{3}{x}\right)\:+{cosx}\right)\right. \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:{cosx}\:+\frac{\mathrm{1}}{\mathrm{4}}\:{cos}\left(\mathrm{3}{x}\right) \\ $$$${S}=\mathrm{6}\sum_{{k}=\mathrm{1}} ^{\mathrm{100}} {cos}\left(\mathrm{2}\frac{{k}\pi}{\mathrm{3}}\right)\:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{\mathrm{100}} {cos}\left(\mathrm{2}{k}\pi\right) \\ $$$${S}=\:\mathrm{200}\:+\mathrm{6}\:{Re}\left(\sum_{{k}=\mathrm{1}} ^{\mathrm{100}} \:{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{3}}} \:\right)\:\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{\mathrm{100}} {e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}{k}} \:=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{100}} \:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}{k}} \:−\mathrm{1}\:\:=\:\frac{\mathrm{1}−\left({e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\right)^{\mathrm{101}} }{\mathrm{1}−{e}^{{i}^{\frac{\mathrm{2}\pi}{\mathrm{3}}} } }\:\:−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\:{e}^{{i}^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \left(\mathrm{99}+\mathrm{2}\right)} }{\mathrm{1}−{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} }\:−\mathrm{1}=\:\frac{\mathrm{1}−{j}^{\mathrm{2}} }{\mathrm{1}−{j}}\:−\mathrm{1}=\:\mathrm{1}+{j}\:−\mathrm{1}={j}\:\:{and}\: \\ $$$${Re}\left(\Sigma\left(..\right)\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow\:{S}=\mathrm{200}\:−\mathrm{3}=\mathrm{197}. \\ $$$$ \\ $$

Commented by abdo imad last updated on 01/Jan/18

S=200−3= 197 .

$${S}=\mathrm{200}−\mathrm{3}=\:\mathrm{197}\:. \\ $$

Answered by AHSoomro last updated on 01/Jan/18

x^2 +x+1=0 x=((−1±(√(1−4)))/2)=((−1±i(√3))/2) =ω,ω^2 Let x=ω (ω+(1/ω))^3 +(ω^2 +(1/ω^2 ))^ω +...+(ω^(100) +(1/ω^(100) ))^3 (ω+ω^(−1) )^3 +(ω^2 +ω^(−2) )^3 +...(ω^(100) +ω^(−100) ) ω^n +ω^(−n) =? n may be one of 3k,3k+1,3k+2 ^• n=3k⇒ω^n +ω^(−n) =ω^(3k) +ω^(−3k) =(ω^3 )^k +(ω^3 )^(−k) =1^k +1^(−k) =2 ^• n=3k+1⇒ω^(3k+1) +ω^(−(3k+1)) =(ω^3 )^k ω+(ω^3 )^(−k) ω^(−1) =ω+ω^(−1) =ω+ω^2 =−1 ^• n=3k+2⇒ω^(3k+2) +ω^(−(3k+2)) =ω^2 +ω^(−2) =ω^2 +ω=−1 (ω^(3k) +(1/ω^(3k) ))^3 33 times 2^3 ×33=264 (ω^(3k+1) +(1/ω^(3k+1) ))^3 and (ω^(3k+2) +(1/ω^(3k+2) ))^3 67 times (−1)^3 ×67=−67 2^3 ×33+(−1)^3 ×67=264−67=197

$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:=\omega,\omega^{\mathrm{2}} \\ $$$$\mathrm{Let}\:\mathrm{x}=\omega \\ $$$$\left(\omega+\frac{\mathrm{1}}{\omega}\right)^{\mathrm{3}} +\left(\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\omega^{\mathrm{2}} }\right)^{\omega} +...+\left(\omega^{\mathrm{100}} +\frac{\mathrm{1}}{\omega^{\mathrm{100}} }\right)^{\mathrm{3}} \\ $$$$\left(\omega+\omega^{−\mathrm{1}} \right)^{\mathrm{3}} +\left(\omega^{\mathrm{2}} +\omega^{−\mathrm{2}} \right)^{\mathrm{3}} +...\left(\omega^{\mathrm{100}} +\omega^{−\mathrm{100}} \right) \\ $$$$\omega^{\mathrm{n}} +\omega^{−\mathrm{n}} =? \\ $$$$\mathrm{n}\:\mathrm{may}\:\mathrm{be}\:\mathrm{one}\:\mathrm{of}\:\mathrm{3k},\mathrm{3k}+\mathrm{1},\mathrm{3k}+\mathrm{2} \\ $$$$\:^{\bullet} \mathrm{n}=\mathrm{3k}\Rightarrow\omega^{\mathrm{n}} +\omega^{−\mathrm{n}} =\omega^{\mathrm{3k}} +\omega^{−\mathrm{3k}} \\ $$$$=\left(\omega^{\mathrm{3}} \right)^{\mathrm{k}} +\left(\omega^{\mathrm{3}} \right)^{−\mathrm{k}} =\mathrm{1}^{\mathrm{k}} +\mathrm{1}^{−\mathrm{k}} =\mathrm{2} \\ $$$$\:^{\bullet} \mathrm{n}=\mathrm{3k}+\mathrm{1}\Rightarrow\omega^{\mathrm{3k}+\mathrm{1}} +\omega^{−\left(\mathrm{3k}+\mathrm{1}\right)} \\ $$$$\:\:=\left(\omega^{\mathrm{3}} \right)^{\mathrm{k}} \omega+\left(\omega^{\mathrm{3}} \right)^{−\mathrm{k}} \omega^{−\mathrm{1}} =\omega+\omega^{−\mathrm{1}} \\ $$$$\:\:=\omega+\omega^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:^{\bullet} \mathrm{n}=\mathrm{3k}+\mathrm{2}\Rightarrow\omega^{\mathrm{3k}+\mathrm{2}} +\omega^{−\left(\mathrm{3k}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:=\omega^{\mathrm{2}} +\omega^{−\mathrm{2}} =\omega^{\mathrm{2}} +\omega=−\mathrm{1} \\ $$$$\left(\omega^{\mathrm{3k}} +\frac{\mathrm{1}}{\omega^{\mathrm{3k}} }\right)^{\mathrm{3}} \:\:\:\mathrm{33}\:\mathrm{times} \\ $$$$\mathrm{2}^{\mathrm{3}} ×\mathrm{33}=\mathrm{264} \\ $$$$\left(\omega^{\mathrm{3k}+\mathrm{1}} +\frac{\mathrm{1}}{\omega^{\mathrm{3k}+\mathrm{1}} }\right)^{\mathrm{3}} \:\:\mathrm{and}\:\:\left(\omega^{\mathrm{3k}+\mathrm{2}} +\frac{\mathrm{1}}{\omega^{\mathrm{3k}+\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{67}\:\mathrm{times} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} ×\mathrm{67}=−\mathrm{67} \\ $$$$\mathrm{2}^{\mathrm{3}} ×\mathrm{33}+\left(−\mathrm{1}\right)^{\mathrm{3}} ×\mathrm{67}=\mathrm{264}−\mathrm{67}=\mathrm{197} \\ $$

Commented by math solver last updated on 01/Jan/18

thanks!

$${thanks}! \\ $$

Commented by abdo imad last updated on 01/Jan/18

you must see that w^n +w^(−n) =w^n +w^−^n = 2Re( w^n ) = 2cos(((2nπ)/3)).

$${you}\:{must}\:{see}\:{that}\:{w}^{{n}} +{w}^{−{n}} ={w}^{{n}} +{w}^{−^{{n}} } \:=\:\mathrm{2}{Re}\left(\:{w}^{{n}} \right)\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right). \\ $$

Commented by Rasheed.Sindhi last updated on 01/Jan/18

I think Mr AHSoomro has solved without going into polar coordinates.

$${I}\:{think}\:{Mr}\:{AHSoomro}\:{has}\:{solved}\:{without} \\ $$$${going}\:{into}\:{polar}\:{coordinates}. \\ $$

Commented by abdo imad last updated on 02/Jan/18

yes it seems...

$${yes}\:{it}\:{seems}... \\ $$

Answered by ajfour last updated on 01/Jan/18

(ω^( r) +(1/ω^( r) ) )^3 = ω^( 3r) +3ω^( r) +(3/ω^( r) )+(1/ω^( 3r) ) =1+3(ω^( r) +(1/ω^( r) ))+1 Σ_(r=1) ^(100) (ω^( r) +(1/ω^( r) ))^3 =200+3Σ_(r=1) ^(100) ω^( r) +3Σ_(r=1) ^(100) (1/ω^( r) ) =200+3×33(ω+ω^2 +1)+3ω +3×33(ω^2 +1+ω)+(3/ω) =200+0+3ω+0+(3/ω) =200+3(ω+ω^2 ) =200−3 = 197 .

$$\left(\omega^{\:{r}} +\frac{\mathrm{1}}{\omega^{\:{r}} }\:\right)^{\mathrm{3}} =\:\omega^{\:\mathrm{3}{r}} +\mathrm{3}\omega^{\:{r}} +\frac{\mathrm{3}}{\omega^{\:{r}} }+\frac{\mathrm{1}}{\omega^{\:\mathrm{3}{r}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{3}\left(\omega^{\:{r}} +\frac{\mathrm{1}}{\omega^{\:{r}} }\right)+\mathrm{1} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\left(\omega^{\:{r}} +\frac{\mathrm{1}}{\omega^{\:{r}} }\right)^{\mathrm{3}} =\mathrm{200}+\mathrm{3}\underset{{r}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\omega^{\:{r}} +\mathrm{3}\underset{{r}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{\mathrm{1}}{\omega^{\:{r}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{200}+\mathrm{3}×\mathrm{33}\left(\omega+\omega^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{3}\omega \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}×\mathrm{33}\left(\omega^{\mathrm{2}} +\mathrm{1}+\omega\right)+\frac{\mathrm{3}}{\omega} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{200}+\mathrm{0}+\mathrm{3}\omega+\mathrm{0}+\frac{\mathrm{3}}{\omega} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{200}+\mathrm{3}\left(\omega+\omega^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{200}−\mathrm{3}\:=\:\mathrm{197}\:. \\ $$

Commented by math solver last updated on 01/Jan/18

thank u sir!

$${thank}\:{u}\:{sir}! \\ $$$$ \\ $$

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