Question-208441

Question Number 208441 by efronzo1 last updated on 16/Jun/24

Answered by Etimbuk last updated on 16/Jun/24

x^3 −y^3 =0 x^(3 ) = y^3 ⇒ x = y the stated fraction will be ((x^(2016) + (x^2 )^(1008) + y^(2016) )/((2x)^(2016) )) = ((2x^(2016) + x^(2016) )/(2^(2016) x^(2016) )) ((3x^(2016) )/(2^(2016) x^(2016) )) = (3/2^(2016) )

$$\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{3}\:} =\:\mathrm{y}^{\mathrm{3}} \:\Rightarrow\:\mathrm{x}\:=\:\mathrm{y} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{stated}\:\mathrm{fraction}\:\mathrm{will}\:\mathrm{be} \\ $$$$\frac{\mathrm{x}^{\mathrm{2016}} \:+\:\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{1008}} \:+\:\mathrm{y}^{\mathrm{2016}} }{\left(\mathrm{2x}\right)^{\mathrm{2016}} }\:=\:\frac{\mathrm{2x}^{\mathrm{2016}} +\:\mathrm{x}^{\mathrm{2016}} }{\mathrm{2}^{\mathrm{2016}} \mathrm{x}^{\mathrm{2016}} } \\ $$$$\frac{\mathrm{3x}^{\mathrm{2016}} }{\mathrm{2}^{\mathrm{2016}} \mathrm{x}^{\mathrm{2016}} }\:=\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{2016}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by efronzo1 last updated on 16/Jun/24

$$\:\:\underline{\:} \\ $$

Answered by Berbere last updated on 16/Jun/24

if y=0⇒x=0 y≠0 ⇒((x/y))^2 +((x/y))+1=0z=(x/y) ⇒;z≠1z^2 +z+1=0⇔z^3 =1;z+1=−z^2 ((z^(2016) +z^(1008) +1)/((1+z)^(2016) ))=(((z^(3)^(672) ) +(z^3 )^(336) +1)/((−z^2 )^(2016) ))=(3/1)=3

$${if}\:{y}=\mathrm{0}\Rightarrow{x}=\mathrm{0} \\ $$$${y}\neq\mathrm{0}\:\Rightarrow\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{y}}\right)+\mathrm{1}=\mathrm{0}{z}=\frac{{x}}{{y}} \\ $$$$\Rightarrow;{z}\neq\mathrm{1}{z}^{\mathrm{2}} +{z}+\mathrm{1}=\mathrm{0}\Leftrightarrow{z}^{\mathrm{3}} =\mathrm{1};{z}+\mathrm{1}=−{z}^{\mathrm{2}} \\ $$$$\frac{{z}^{\mathrm{2016}} +{z}^{\mathrm{1008}} +\mathrm{1}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2016}} }=\frac{\left({z}^{\left.\mathrm{3}\right)^{\mathrm{672}} } +\left({z}^{\mathrm{3}} \right)^{\mathrm{336}} +\mathrm{1}\right.}{\left(−{z}^{\mathrm{2}} \right)^{\mathrm{2016}} }=\frac{\mathrm{3}}{\mathrm{1}}=\mathrm{3} \\ $$$$ \\ $$

Answered by Frix last updated on 16/Jun/24

x^2 +xy+y^2 =0 y=px∧x≠0 ⇒l p^2 +p+1=0 ((p^(2016) +p^(1008) +1)/((p+1)^(2016) )) p=e^(±i((2π)/3)) ⇒ p+1=e^(±i(π/3)) p^(3n) =1∧(p+1)^(6n) =1 ⇒ ((p^(2016) +p^(1008) +1)/((p+1)^(2016) ))=3

$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}={px}\wedge{x}\neq\mathrm{0}\:\Rightarrow{l} \\ $$$${p}^{\mathrm{2}} +{p}+\mathrm{1}=\mathrm{0} \\ $$$$\frac{{p}^{\mathrm{2016}} +{p}^{\mathrm{1008}} +\mathrm{1}}{\left({p}+\mathrm{1}\right)^{\mathrm{2016}} } \\ $$$${p}=\mathrm{e}^{\pm\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow\:{p}+\mathrm{1}=\mathrm{e}^{\pm\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$${p}^{\mathrm{3}{n}} =\mathrm{1}\wedge\left({p}+\mathrm{1}\right)^{\mathrm{6}{n}} =\mathrm{1}\:\Rightarrow\:\frac{{p}^{\mathrm{2016}} +{p}^{\mathrm{1008}} +\mathrm{1}}{\left({p}+\mathrm{1}\right)^{\mathrm{2016}} }=\mathrm{3} \\ $$

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