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Question-7246




Question Number 7246 by Tawakalitu. last updated on 18/Aug/16
Answered by Yozzia last updated on 19/Aug/16
Define the number k such that   k=x^(n+3) +y^(n+3) +z^(n+3) −xyz(x^n +y^n +z^n )         −(1/2)(x^2 +y^2 +z^2 )(x^(n+1) +y^(n+1) +z^(n+1) )  x,y,z,n∈C and x+y+z=0.  −−−−−−−−−−−−−−−−−−−−−−−−  −(1/2)(x^2 +y^2 +z^2 )(x^(n+1) +y^(n+1) +z^(n+1) )  =−(1/2){x^(n+3) +y^(n+3) +z^(n+3) +x^(n+1) (y^2 +z^2 )         +y^(n+1) (x^2 +z^2 )+z^(n+1) (x^2 +y^2 )}    Also, −xyz(x^n +y^n +z^n )=−x^(n+1) yz−xy^(n+1) z−xyz^(n+1)     ∴ k=(1/2)(x^(n+3) +y^(n+3) +z^(n+3) )−x^(n+1) yz−xy^(n+1) z−xyz^(n+1)               −(1/2){x^(n+1) (y^2 +z^2 )+y^(n+1) (x^2 +z^2 )+z^(n+1) (x^2 +y^2 )}  k=(1/2){x^(n+3) +y^(n+3) +z^(n+3) −x^(n+1) (y^2 +z^2 +2zy)−y^(n+1) (x^2 +z^2 +2zx)−z^(n+1) (x^2 +y^2 +2xy)}  k=(1/2){x^(n+3) +y^(n+3) +z^(n+3) −x^(n+1) (y+z)^2 −y^(n+1) (x+z)^2 −z^(n+1) (x+y)^2 }  Since x+y+z=0⇒ x+y=−z, x+z=−y, y+z=−x.  ∴ k=(1/2){x^(n+3) +y^(n+3) +z^(n+3) −x^(n+1) (−x)^2 −y^(n+1) (−y)^2 −z^(n+1) (−z)^2 }   k=(1/2){x^(n+3) +y^(n+3) +z^(n+3) −x^(n+3) −y^(n+3) −z^(n+3) }  k=0, if x+y+z=0.  Thus x^(n+3) +y^(n+3) +z^(n+3) −xyz(x^n +y^n +z^n )−(1/2)(x^2 +y^2 +z^2 )(x^(n+1) +y^(n+1) +z^(n+1) )=0  or   x^(n+3) +y^(n+3) +z^(n+3) =xyz(x^n +y^n +z^n )+(1/2)(x^2 +y^2 +z^2 )(x^(n+1) +y^(n+1) +z^(n+1) )  if x+y+z=0.
$${Define}\:{the}\:{number}\:{k}\:{such}\:{that}\: \\ $$$${k}={x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right) \\ $$$$\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right) \\ $$$${x},{y},{z},{n}\in\mathbb{C}\:{and}\:{x}+{y}+{z}=\mathrm{0}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} +{x}^{{n}+\mathrm{1}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right. \\ $$$$\left.\:\:\:\:\:\:\:+{y}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{z}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right\} \\ $$$$ \\ $$$${Also},\:−{xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right)=−{x}^{{n}+\mathrm{1}} {yz}−{xy}^{{n}+\mathrm{1}} {z}−{xyz}^{{n}+\mathrm{1}} \\ $$$$ \\ $$$$\therefore\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} \right)−{x}^{{n}+\mathrm{1}} {yz}−{xy}^{{n}+\mathrm{1}} {z}−{xyz}^{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{1}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{y}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{z}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right\} \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{1}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{zy}\right)−{y}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{zx}\right)−{z}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}\right)\right\} \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{1}} \left({y}+{z}\right)^{\mathrm{2}} −{y}^{{n}+\mathrm{1}} \left({x}+{z}\right)^{\mathrm{2}} −{z}^{{n}+\mathrm{1}} \left({x}+{y}\right)^{\mathrm{2}} \right\} \\ $$$${Since}\:{x}+{y}+{z}=\mathrm{0}\Rightarrow\:{x}+{y}=−{z},\:{x}+{z}=−{y},\:{y}+{z}=−{x}. \\ $$$$\therefore\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{1}} \left(−{x}\right)^{\mathrm{2}} −{y}^{{n}+\mathrm{1}} \left(−{y}\right)^{\mathrm{2}} −{z}^{{n}+\mathrm{1}} \left(−{z}\right)^{\mathrm{2}} \right\} \\ $$$$\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{3}} −{y}^{{n}+\mathrm{3}} −{z}^{{n}+\mathrm{3}} \right\} \\ $$$${k}=\mathrm{0},\:{if}\:{x}+{y}+{z}=\mathrm{0}. \\ $$$${Thus}\:{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right)=\mathrm{0} \\ $$$${or}\: \\ $$$${x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} ={xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right) \\ $$$${if}\:{x}+{y}+{z}=\mathrm{0}. \\ $$
Commented by Tawakalitu. last updated on 19/Aug/16
Wow thanks so muvh sir. i really appreciate.
$${Wow}\:{thanks}\:{so}\:{muvh}\:{sir}.\:{i}\:{really}\:{appreciate}. \\ $$

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