Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 158124 by HongKing last updated on 31/Oct/21

let 𝛚 be a root of the equation x^4 + (x - 1)^4 + 1 = 0 find 𝛀 = 𝛚^(300) + 𝛚^(301)

$$\mathrm{let}\:\boldsymbol{\omega}\:\mathrm{be}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\left(\mathrm{x}\:-\:\mathrm{1}\right)^{\mathrm{4}} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}\:\:\boldsymbol{\Omega}\:=\:\boldsymbol{\omega}^{\mathrm{300}} \:+\:\boldsymbol{\omega}^{\mathrm{301}} \\ $$

Answered by Rasheed.Sindhi last updated on 31/Oct/21

x^4 + (x - 1)^4 + 1 = 0; 𝛀 = 𝛚^(300) + 𝛚^(301) ? 2x^4 −4x^3 +6x^2 −4x+2=0 x^4 −2x^3 +3x^2 −2x+1=0 x^2 −2x+3−(2/x)+(1/x^2 )=0: x≠0 x^2 +2+(1/x^2 )−2(x+(1/x))_(y) +1=0 y^2 −2y+1=0⇒(y−1)^2 =0⇒y=1 x+(1/x)=1⇒x^2 −x+1=0 x=((1±(√(1−4)))/2)=−((−1∓i(√3))/2)= ∴ ω is negative of cuberoot.Let one cuberoot of unity is c ∴ c^3 =1 𝛀 = 𝛚^(300) + 𝛚^(301) =(−c)^(300) +(−c)^(301) =c^(300) −c^(301) =(c^3 )^(100) −(c^3 )^(100) .c =(1)^(100) −(1)^(100) .c=1−c =1−((−1+i(√3))/2)=((2+1−i(√3))/2)=((3−i(√3))/2) OR =1−((−1−i(√3))/2)=((2+1+i(√3))/2)=((3+i(√3))/2)

$$\mathrm{x}^{\mathrm{4}} \:+\:\left(\mathrm{x}\:-\:\mathrm{1}\right)^{\mathrm{4}} \:+\:\mathrm{1}\:=\:\mathrm{0};\:\:\:\boldsymbol{\Omega}\:=\:\boldsymbol{\omega}^{\mathrm{300}} \:+\:\boldsymbol{\omega}^{\mathrm{301}} ? \\ $$$$\mathrm{2x}^{\mathrm{4}} −\mathrm{4x}^{\mathrm{3}} +\mathrm{6x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{3}−\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{0}:\:\mathrm{x}\neq\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{2}\underset{\mathrm{y}} {\underbrace{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)}}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{2y}+\mathrm{1}=\mathrm{0}\Rightarrow\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{y}=\mathrm{1} \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{1}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}=−\frac{−\mathrm{1}\mp\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}= \\ $$$$\therefore\:\omega\:{is}\:{negative}\:{of}\:{cuberoot}.{Let} \\ $$$${one}\:{cuberoot}\:{of}\:{unity}\:{is}\:{c} \\ $$$$\therefore\:{c}^{\mathrm{3}} =\mathrm{1} \\ $$$$\boldsymbol{\Omega}\:=\:\boldsymbol{\omega}^{\mathrm{300}} \:+\:\boldsymbol{\omega}^{\mathrm{301}} \:\: \\ $$$$\:\:\:\:\:\:=\left(−{c}\right)^{\mathrm{300}} +\left(−{c}\right)^{\mathrm{301}} \\ $$$$\:\:\:\:\:={c}^{\mathrm{300}} −{c}^{\mathrm{301}} =\left({c}^{\mathrm{3}} \right)^{\mathrm{100}} −\left({c}^{\mathrm{3}} \right)^{\mathrm{100}} .{c} \\ $$$$\:\:\:\:=\left(\mathrm{1}\right)^{\mathrm{100}} −\left(\mathrm{1}\right)^{\mathrm{100}} .{c}=\mathrm{1}−{c} \\ $$$$\:\:\:=\mathrm{1}−\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{OR}\: \\ $$$$\:\:=\mathrm{1}−\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 04/Nov/21

ω^3 =−1 is correct. Ω=((3±i(√3))/2) is correct. but Ω≠0

$$\omega^{\mathrm{3}} =−\mathrm{1}\:{is}\:{correct}. \\ $$$$\Omega=\frac{\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{is}\:{correct}. \\ $$$${but}\:\Omega\neq\mathrm{0} \\ $$

Commented by HongKing last updated on 31/Oct/21

thank you dear Ser but 𝛚^3 =-1 ⇒ 𝛀=0

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Ser} \\ $$$$\mathrm{but}\:\boldsymbol{\omega}^{\mathrm{3}} =-\mathrm{1}\:\Rightarrow\:\boldsymbol{\Omega}=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 31/Oct/21

𝛀 = 𝛚^(300) + 𝛚^(301) =(ω^3 )^(100) +(ω^3 )^(100) .ω =(−1)^(100) +(−1)^(100) .ω =1+ω =1+(−c)=1−c=1−ω (OR =1−ω^2 )

$$\:\:\boldsymbol{\Omega}\:=\:\boldsymbol{\omega}^{\mathrm{300}} \:+\:\boldsymbol{\omega}^{\mathrm{301}} \\ $$$$\:\:\:\:\:\:\:=\left(\omega^{\mathrm{3}} \right)^{\mathrm{100}} +\left(\omega^{\mathrm{3}} \right)^{\mathrm{100}} .\omega \\ $$$$\:\:\:\:\:=\left(−\mathrm{1}\right)^{\mathrm{100}} +\left(−\mathrm{1}\right)^{\mathrm{100}} .\omega \\ $$$$\:\:\:\:\:=\mathrm{1}+\omega \\ $$$$\:\:\:\:=\mathrm{1}+\left(−{c}\right)=\mathrm{1}−{c}=\mathrm{1}−\omega \\ $$$$\left(\mathrm{OR}\:=\mathrm{1}−\omega^{\mathrm{2}} \right) \\ $$

Commented by Rasheed.Sindhi last updated on 01/Nov/21

Agree?

$$\mathrm{Agree}? \\ $$

Commented by HongKing last updated on 04/Nov/21

answer: 0 dear Ser

$$\mathrm{answer}:\:\mathrm{0}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Commented by Rasheed.Sindhi last updated on 04/Nov/21

Answer you have may be wrong.The above comment is proof.

$${Answer}\:{you}\:{have}\:{may}\:{be}\:{wrong}.{The} \\ $$$${above}\:{comment}\:{is}\:{proof}. \\ $$

Commented by Rasheed.Sindhi last updated on 05/Nov/21

Thanks a Lot Sir!

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{a}\:\mathbb{L}\mathrm{ot}\:\mathbb{S}\mathrm{ir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com