Question Number 123824 by Snail last updated on 28/Nov/20
$${Show}\:{that}\:{for}\:{all}\:{real}\:{numbers}\:\left({x}/{y}/{z}\right)\:{satisfying}\:\: \\ $$$${x}+{y}+{z}=\mathrm{0}\:{and}\:{xy}\:+{yz}+{zx}=−\mathrm{3}\:\:\:{the}\:{value}\:{of} \\ $$$${expression}\:{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}\:+{z}^{\mathrm{3}} {x}\:\:\:{is}\:{a}\:{constant} \\ $$
Commented by Snail last updated on 28/Nov/20
$${I}\:{think}\:{u}\:{have}\:{done}\:{a}\:{mistake} \\ $$$${u}\:{have}\:{written}\: \\ $$$${x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {x}={A}\left({let}\right)\:{and}\:{xy}^{\mathrm{3}} +{yz}^{\mathrm{3}} +{x}^{\mathrm{3}} {z}={B}\left({let}\right) \\ $$$${thenu}\:{have}\:{done}\: \\ $$$${A}=−\mathrm{18}−{B}\Rightarrow\left({a}\right) \\ $$$${and}\:{then}\:{B}=−\mathrm{18}−{A} \\ $$$${But}\:{that}\:{does}\:{not}\:{imply}\:{A}={B} \\ $$
Answered by MJS_new last updated on 28/Nov/20
$${z}=−\left({x}+{y}\right) \\ $$$$\Rightarrow \\ $$$${xy}+{yz}+{zx}=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)\wedge{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {y}=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} =−\mathrm{9} \\ $$
Commented by Dwaipayan Shikari last updated on 29/Nov/20
$${z}=−\left({x}+{y}\right) \\ $$$${xy}+{yz}+{xz}={xy}+{y}\left(−{x}−{y}\right)−\left({x}+{y}\right){x}=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {y}\: \\ $$$$={x}^{\mathrm{3}} {y}−{y}^{\mathrm{3}} \left({x}+{y}\right)−\left({x}+{y}\right)^{\mathrm{3}} {y} \\ $$$$=−\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} =−\mathrm{9} \\ $$
Commented by Snail last updated on 29/Nov/20
$${I}\:{can}'{t}\:{understand}\:{the}\:{process} \\ $$