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Question Number 111624 by mathdave last updated on 04/Sep/20

if x is a cube root of a unity prove that (1−x)^6 =−27

$${if}\:{x}\:{is}\:{a}\:{cube}\:{root}\:{of}\:{a}\:{unity} \\ $$$${prove}\:{that}\: \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =−\mathrm{27} \\ $$

Answered by Her_Majesty last updated on 04/Sep/20

(1−ω)^6 =ω^6 −6ω^5 +15ω^4 −20ω^3 +15ω^2 −6ω+1= =1 −6ω^2 +15ω −20 +15ω^2 −6ω+1= =9(ω+ω^2 )−18=−27

$$\left(\mathrm{1}−\omega\right)^{\mathrm{6}} =\omega^{\mathrm{6}} −\mathrm{6}\omega^{\mathrm{5}} +\mathrm{15}\omega^{\mathrm{4}} −\mathrm{20}\omega^{\mathrm{3}} +\mathrm{15}\omega^{\mathrm{2}} −\mathrm{6}\omega+\mathrm{1}= \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{1}\:\:−\mathrm{6}\omega^{\mathrm{2}} +\mathrm{15}\omega\:\:−\mathrm{20}\:\:\:\:\:\:+\mathrm{15}\omega^{\mathrm{2}} −\mathrm{6}\omega+\mathrm{1}= \\ $$$$=\mathrm{9}\left(\omega+\omega^{\mathrm{2}} \right)−\mathrm{18}=−\mathrm{27} \\ $$

Answered by $@y@m last updated on 04/Sep/20

(1−x)^6 ={(1−x)^3 }^2 ={1−x^3 −3x(1−x)}^2 ={1−1−3x(1−x)}^2 =9x^2 (1−2x+x^2 ) =9(x^2 −2x^3 +x^4 ) =9(x^2 −2+x) =9(−2−1) =−27

$$\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =\left\{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \right\}^{\mathrm{2}} \\ $$$$=\left\{\mathrm{1}−{x}^{\mathrm{3}} −\mathrm{3}{x}\left(\mathrm{1}−{x}\right)\right\}^{\mathrm{2}} \\ $$$$=\left\{\mathrm{1}−\mathrm{1}−\mathrm{3}{x}\left(\mathrm{1}−{x}\right)\right\}^{\mathrm{2}} \\ $$$$=\mathrm{9}{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{9}\left({x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{4}} \right) \\ $$$$=\mathrm{9}\left({x}^{\mathrm{2}} −\mathrm{2}+{x}\right) \\ $$$$=\mathrm{9}\left(−\mathrm{2}−\mathrm{1}\right) \\ $$$$=−\mathrm{27} \\ $$

Commented by Rasheed.Sindhi last updated on 04/Sep/20

Cool!

$${Cool}! \\ $$

Commented by $@y@m last updated on 04/Sep/20

Thanks

$${Thanks} \\ $$

Answered by mathdave last updated on 04/Sep/20

solution let y^3 =1 y=(1)^(1/3) =(cos(0)+jsin(0))^(1/3) to find the roots let y=(cos2nπ+jsin2nπ)^(1/3) y=(cos((2nπ)/3)+jsin((2nπ)/3)) putting n=0,1,2 which are the roots of the unity at y_n =y_0 =1..........(1) y_1 =[cos((2π)/3)+jsin((2π)/3)]^1 =x ..........(2),at y_2 =[cos((4π)/3)+jsin((4π)/3)]^2 =x^2 ........(3) adding (1),(2), and (3) 1+x+x^2 =1+cos((2π)/3)+jsin((2π)/3)+cos((4π)/3)+jsin((4π)/3) 1+x+x^2 =1+cos(π−(π/3))+jsin(π−(π/3))+cos(π+(π/3))+jsin(π+(π/3)) 1+x+x^2 =1−cos(π/3)+jsin(π/3)−cos((.π)/3)−jsin(π/3) 1+x+x^2 =1−2cos(π/3)=1−2((1/2))=0 1+x+x^2 =0 1+x^2 =−x.........(1) but (1−x)^6 =[(1−x)^2 ]^3 (1−x)^6 =(1−2x+x^2 )^3 =[(1+x^2 )−2x]^3 (1−x)^6 =(−x−2x)^3 =(−3x)^3 (1−x)^6 =−27x^3 but x^3 =1 ∵(1−x)^6 =−27 Q.E.D

$${solution}\: \\ $$$${let}\:{y}^{\mathrm{3}} =\mathrm{1} \\ $$$${y}=\left(\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(\mathrm{cos}\left(\mathrm{0}\right)+{j}\mathrm{sin}\left(\mathrm{0}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:{to}\:{find}\:{the}\:{roots}\:{let} \\ $$$${y}=\left(\mathrm{cos2}{n}\pi+{j}\mathrm{sin2}{n}\pi\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${y}=\left(\mathrm{cos}\frac{\mathrm{2}{n}\pi}{\mathrm{3}}+{j}\mathrm{sin}\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)\:\:\:{putting}\:{n}=\mathrm{0},\mathrm{1},\mathrm{2}\:{which}\:{are}\:{the}\:{roots}\:{of}\:{the}\:{unity} \\ $$$${at}\:\:{y}_{{n}} ={y}_{\mathrm{0}} =\mathrm{1}..........\left(\mathrm{1}\right) \\ $$$${y}_{\mathrm{1}} =\left[\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}+{j}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{3}}\right]^{\mathrm{1}} ={x}\:\:..........\left(\mathrm{2}\right),{at}\: \\ $$$${y}_{\mathrm{2}} =\left[\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{3}}+{j}\mathrm{sin}\frac{\mathrm{4}\pi}{\mathrm{3}}\right]^{\mathrm{2}} ={x}^{\mathrm{2}} ........\left(\mathrm{3}\right) \\ $$$${adding}\:\left(\mathrm{1}\right),\left(\mathrm{2}\right),\:{and}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}+{j}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{3}}+{j}\mathrm{sin}\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{cos}\left(\pi−\frac{\pi}{\mathrm{3}}\right)+{j}\mathrm{sin}\left(\pi−\frac{\pi}{\mathrm{3}}\right)+\mathrm{cos}\left(\pi+\frac{\pi}{\mathrm{3}}\right)+{j}\mathrm{sin}\left(\pi+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{1}−\mathrm{cos}\frac{\pi}{\mathrm{3}}+{j}\mathrm{sin}\frac{\pi}{\mathrm{3}}−\mathrm{cos}\frac{.\pi}{\mathrm{3}}−{j}\mathrm{sin}\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{1}−\mathrm{2cos}\frac{\pi}{\mathrm{3}}=\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} =−{x}.........\left(\mathrm{1}\right) \\ $$$${but}\:\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =\left[\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \right]^{\mathrm{3}} \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =\left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} =\left[\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}\right]^{\mathrm{3}} \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =\left(−{x}−\mathrm{2}{x}\right)^{\mathrm{3}} =\left(−\mathrm{3}{x}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =−\mathrm{27}{x}^{\mathrm{3}} \\ $$$${but}\:\:{x}^{\mathrm{3}} =\mathrm{1} \\ $$$$\because\left(\mathrm{1}−{x}\right)^{\mathrm{6}} =−\mathrm{27}\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

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