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x-y-1-x-5-xy-y-5-x-2-y-2-faind-the-volue-of-x-2022-xy-y-2022-




Question Number 174029 by mathlove last updated on 23/Jul/22
x+y=1  x^5 +xy+y^5 =x^2 y^(2      )   faind the volue of   x^(2022) +xy+y^(2022) =?
$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{5}} +{xy}+{y}^{\mathrm{5}} ={x}^{\mathrm{2}} {y}^{\mathrm{2}\:\:\:\:\:\:} \\ $$$${faind}\:{the}\:{volue}\:{of}\:\:\:{x}^{\mathrm{2022}} +{xy}+{y}^{\mathrm{2022}} =? \\ $$
Commented by MJS_new last updated on 23/Jul/22
y=1−x  4x^4 −8x^3 +8x^2 −4x+1=0  (2x^2 −2x+1)^2 =0  x=(1/2)±(1/2)i ∧ y=conj (x) ⇒ x^n +y^n =0∧xy=(1/2)  answer is (1/2)
$${y}=\mathrm{1}−{x} \\ $$$$\mathrm{4}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\:\wedge\:{y}=\mathrm{conj}\:\left({x}\right)\:\Rightarrow\:{x}^{{n}} +{y}^{{n}} =\mathrm{0}\wedge{xy}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 23/Jul/22
x^2 +y^2 +2xy=1  x^2 +y^2 =1−2xy    (x^2 +y^2 )(x+y)=1−2xy  x^3 +y^3 +xy(x+y)=1−2xy  x^3 +y^3 =1−3xy    x^4 +y^4 +2(xy)^2 =1−4xy+4(xy)^2   x^4 +y^4 =1−4xy+2(xy)^2     (x^4 +y^4 )(x+y)=1−4xy+2(xy)^2   x^5 +y^5 +xy(x^3 +y^3 )=1−4xy+2(xy)^2   x^5 +y^5 +xy(1−3xy)=1−4xy+2(xy)^2   x^5 +y^5 =1−5xy+5(xy)^2     (xy)^2 −xy=1−5xy+5(xy)^2   0=1−4xy+4(xy)^2   (2xy−1)^2 =0  ⇒xy=(1/2)  x,y are roots of  z^2 −z+(1/2)=0  z=(1/2)(1±i)=(1/( (√2)))((1/( (√2)))±(i/( (√2))))     =(1/( (√2)))[cos (±(π/4))+i sin (±(π/4))]    z_1 ^(2022) =(1/2^(1011) )[cos (((2022π)/4))+i sin (((2022π)/4))]  z_1 ^(2022) =(1/2^(1011) )[cos (505π+(π/2))+i sin (505π+(π/2))]  z_1 ^(2022) =−(i/2^(1011) )  z_2 ^(2022) =(1/2^(1011) )[cos (−((2022π)/4))+i sin (−((2022π)/4))]  z_2 ^(2022) =(1/2^(1011) )[cos (−505π−(π/2))+i sin (−505π−(π/2))]  z_2 ^(2022) =(i/2^(1011) )  ⇒x^(2022) +y^(2022) =z_1 ^(2022) +z_2 ^(2022) =0  ⇒x^(2022) +xy+y^(2022) =(1/2) ✓
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{xy} \\ $$$$ \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}+{y}\right)=\mathrm{1}−\mathrm{2}{xy} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{xy}\left({x}+{y}\right)=\mathrm{1}−\mathrm{2}{xy} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{1}−\mathrm{3}{xy} \\ $$$$ \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\mathrm{2}\left({xy}\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{4}{xy}+\mathrm{4}\left({xy}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{1}−\mathrm{4}{xy}+\mathrm{2}\left({xy}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}+{y}\right)=\mathrm{1}−\mathrm{4}{xy}+\mathrm{2}\left({xy}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{xy}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)=\mathrm{1}−\mathrm{4}{xy}+\mathrm{2}\left({xy}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{xy}\left(\mathrm{1}−\mathrm{3}{xy}\right)=\mathrm{1}−\mathrm{4}{xy}+\mathrm{2}\left({xy}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{1}−\mathrm{5}{xy}+\mathrm{5}\left({xy}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left({xy}\right)^{\mathrm{2}} −{xy}=\mathrm{1}−\mathrm{5}{xy}+\mathrm{5}\left({xy}\right)^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{1}−\mathrm{4}{xy}+\mathrm{4}\left({xy}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{xy}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{xy}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x},{y}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −{z}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm{i}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\pm\frac{{i}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{cos}\:\left(\pm\frac{\pi}{\mathrm{4}}\right)+{i}\:\mathrm{sin}\:\left(\pm\frac{\pi}{\mathrm{4}}\right)\right] \\ $$$$ \\ $$$${z}_{\mathrm{1}} ^{\mathrm{2022}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1011}} }\left[\mathrm{cos}\:\left(\frac{\mathrm{2022}\pi}{\mathrm{4}}\right)+{i}\:\mathrm{sin}\:\left(\frac{\mathrm{2022}\pi}{\mathrm{4}}\right)\right] \\ $$$${z}_{\mathrm{1}} ^{\mathrm{2022}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1011}} }\left[\mathrm{cos}\:\left(\mathrm{505}\pi+\frac{\pi}{\mathrm{2}}\right)+{i}\:\mathrm{sin}\:\left(\mathrm{505}\pi+\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$${z}_{\mathrm{1}} ^{\mathrm{2022}} =−\frac{{i}}{\mathrm{2}^{\mathrm{1011}} } \\ $$$${z}_{\mathrm{2}} ^{\mathrm{2022}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1011}} }\left[\mathrm{cos}\:\left(−\frac{\mathrm{2022}\pi}{\mathrm{4}}\right)+{i}\:\mathrm{sin}\:\left(−\frac{\mathrm{2022}\pi}{\mathrm{4}}\right)\right] \\ $$$${z}_{\mathrm{2}} ^{\mathrm{2022}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1011}} }\left[\mathrm{cos}\:\left(−\mathrm{505}\pi−\frac{\pi}{\mathrm{2}}\right)+{i}\:\mathrm{sin}\:\left(−\mathrm{505}\pi−\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$${z}_{\mathrm{2}} ^{\mathrm{2022}} =\frac{{i}}{\mathrm{2}^{\mathrm{1011}} } \\ $$$$\Rightarrow{x}^{\mathrm{2022}} +{y}^{\mathrm{2022}} ={z}_{\mathrm{1}} ^{\mathrm{2022}} +{z}_{\mathrm{2}} ^{\mathrm{2022}} =\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2022}} +{xy}+{y}^{\mathrm{2022}} =\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
Commented by behi834171 last updated on 23/Jul/22
great sir mrW.
$${great}\:{sir}\:{mrW}. \\ $$
Commented by Tawa11 last updated on 23/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mathlove last updated on 23/Jul/22
thanks mr W
$${thanks}\:{mr}\:{W} \\ $$

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