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Question Number 123824 by Snail last updated on 28/Nov/20

Show that for all real numbers (x/y/z) satisfying    x+y+z=0 and xy +yz+zx=−3   the value of  expression x^3 y+y^3 z +z^3 x   is a constant

$${Show}\:{that}\:{for}\:{all}\:{real}\:{numbers}\:\left({x}/{y}/{z}\right)\:{satisfying}\:\: \\ $$$${x}+{y}+{z}=\mathrm{0}\:{and}\:{xy}\:+{yz}+{zx}=−\mathrm{3}\:\:\:{the}\:{value}\:{of} \\ $$$${expression}\:{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}\:+{z}^{\mathrm{3}} {x}\:\:\:{is}\:{a}\:{constant} \\ $$

Commented by Snail last updated on 28/Nov/20

I think u have done a mistake  u have written   x^3 y+y^3 z+z^3 x=A(let) and xy^3 +yz^3 +x^3 z=B(let)  thenu have done   A=−18−B⇒(a)  and then B=−18−A  But that does not imply A=B

$${I}\:{think}\:{u}\:{have}\:{done}\:{a}\:{mistake} \\ $$$${u}\:{have}\:{written}\: \\ $$$${x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {x}={A}\left({let}\right)\:{and}\:{xy}^{\mathrm{3}} +{yz}^{\mathrm{3}} +{x}^{\mathrm{3}} {z}={B}\left({let}\right) \\ $$$${thenu}\:{have}\:{done}\: \\ $$$${A}=−\mathrm{18}−{B}\Rightarrow\left({a}\right) \\ $$$${and}\:{then}\:{B}=−\mathrm{18}−{A} \\ $$$${But}\:{that}\:{does}\:{not}\:{imply}\:{A}={B} \\ $$

Answered by MJS_new last updated on 28/Nov/20

z=−(x+y)  ⇒  xy+yz+zx=−(x^2 +xy+y^2 )∧x^3 y+y^3 z+z^3 y=−(x^2 +xy+y^2 )^2 =−9

$${z}=−\left({x}+{y}\right) \\ $$$$\Rightarrow \\ $$$${xy}+{yz}+{zx}=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)\wedge{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {y}=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} =−\mathrm{9} \\ $$

Commented by Dwaipayan Shikari last updated on 29/Nov/20

z=−(x+y)  xy+yz+xz=xy+y(−x−y)−(x+y)x=−(x^2 +xy+y^2 )  x^3 y+y^3 z+z^3 y   =x^3 y−y^3 (x+y)−(x+y)^3 y  =−(x^2 +xy+y^2 )^2 =−9

$${z}=−\left({x}+{y}\right) \\ $$$${xy}+{yz}+{xz}={xy}+{y}\left(−{x}−{y}\right)−\left({x}+{y}\right){x}=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {y}\: \\ $$$$={x}^{\mathrm{3}} {y}−{y}^{\mathrm{3}} \left({x}+{y}\right)−\left({x}+{y}\right)^{\mathrm{3}} {y} \\ $$$$=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} =−\mathrm{9} \\ $$

Commented by Snail last updated on 29/Nov/20

I can′t understand the process

$${I}\:{can}'{t}\:{understand}\:{the}\:{process} \\ $$

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