Question-143899

Question Number 143899 by liberty last updated on 19/Jun/21

Answered by ajfour last updated on 19/Jun/21

let r^(10) =z t=z^4 −z^3 +z^2 −z+1 r^5 while ((r^5 −1)/(r−1))=0 let r=cos 72^° +isin 72° ⇒ but anyway r^5 =1 z=r^(10) =1 so t=r^(40) −r^(30) +r^(20) −r^(10) +1 =z^4 −z^3 +z^2 −z+1 =1−1+1−1+1 =1

$${let}\:\:{r}^{\mathrm{10}} ={z} \\ $$$${t}={z}^{\mathrm{4}} −{z}^{\mathrm{3}} +{z}^{\mathrm{2}} −{z}+\mathrm{1} \\ $$$${r}^{\mathrm{5}} \\ $$$${while}\:\:\:\frac{{r}^{\mathrm{5}} −\mathrm{1}}{{r}−\mathrm{1}}=\mathrm{0} \\ $$$${let}\:{r}=\mathrm{cos}\:\mathrm{72}^{°} +{i}\mathrm{sin}\:\mathrm{72}° \\ $$$$\Rightarrow\:{but}\:{anyway}\:{r}^{\mathrm{5}} =\mathrm{1} \\ $$$${z}={r}^{\mathrm{10}} =\mathrm{1} \\ $$$${so}\:{t}={r}^{\mathrm{40}} −{r}^{\mathrm{30}} +{r}^{\mathrm{20}} −{r}^{\mathrm{10}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:={z}^{\mathrm{4}} −{z}^{\mathrm{3}} +{z}^{\mathrm{2}} −{z}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{1} \\ $$

Answered by MJS_new last updated on 19/Jun/21

(x^4 −x^3 +x^2 −x+1)(x+1)=x^5 +1=0 ⇔ x^5 =−1 ⇒ x^(10) =x^(20) =x^(30) =x^(40) =1 ⇒ answer is 1

$$\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)={x}^{\mathrm{5}} +\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:{x}^{\mathrm{5}} =−\mathrm{1} \\ $$$$\Rightarrow\:{x}^{\mathrm{10}} ={x}^{\mathrm{20}} ={x}^{\mathrm{30}} ={x}^{\mathrm{40}} =\mathrm{1}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{1} \\ $$

Answered by mr W last updated on 19/Jun/21

x^4 −x^3 +x^2 −x+1=0 (x+1)(x^4 −x^3 +x^2 −x+1)=0 x^5 +1=0 x^5 =−1 ⇒r^5 =−1 ∧r≠−1 r^(40) −r^(30) +r^(20) −r^(10) +1=((1+r^(50) )/(1+r^(10) ))=((1+(r^5 )^(10) )/(1+(r^5 )^2 )) =((1+(−1)^(10) )/(1+(−1)^2 ))=((1+1)/(1+1))=1

$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{5}} +\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{5}} =−\mathrm{1} \\ $$$$\Rightarrow{r}^{\mathrm{5}} =−\mathrm{1}\:\wedge{r}\neq−\mathrm{1} \\ $$$${r}^{\mathrm{40}} −{r}^{\mathrm{30}} +{r}^{\mathrm{20}} −{r}^{\mathrm{10}} +\mathrm{1}=\frac{\mathrm{1}+{r}^{\mathrm{50}} }{\mathrm{1}+{r}^{\mathrm{10}} }=\frac{\mathrm{1}+\left({r}^{\mathrm{5}} \right)^{\mathrm{10}} }{\mathrm{1}+\left({r}^{\mathrm{5}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{10}} }{\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\mathrm{1} \\ $$

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