Question Number 147187 by mathdanisur last updated on 18/Jul/21
$${if}\:\:\left(\overline {{xyz}}\right)^{\mathrm{2}} \:=\:\left({x}+{y}+{z}\right)^{\mathrm{5}} \:\:{then}\:{find}: \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mid\left({x}+{y}+{z}\right)+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\mid \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jul/21
$${Is}\:\overline {{xyz}}\:{a}\:{digital}\:{representation}\:{of} \\ $$$${decimal}\:{number}? \\ $$
Commented by mathdanisur last updated on 18/Jul/21
$${Yes}\:{Ser} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Jul/21
$$\:\left(\overline {\:{xyz}\:}\right)^{\mathrm{2}} =\left({x}+{y}+{z}\right)^{\mathrm{5}} \\ $$$$\blacktriangleright\:\overline {\:{xyz}\:}\:{is}\:\mathrm{3}-{digit}\:{natural}\:{number} \\ $$$$\blacktriangleright\:\:\left(\overline {\:{xyz}\:}\right)^{\mathrm{2}} \:{is}\:{perfect}\:\mathrm{5}{th}\:{power}\: \\ $$$$\:\:\:\:\:\:\Rightarrow\overline {\:{xyz}\:}\:{is}\:{perfect}\:\:\mathrm{5}{th}\:{power} \\ $$$${The}\:{only}\:{three}\:{digit}\:{number}\:{which} \\ $$$${is}\:{also}\:\mathrm{5}{th}\:{power}\:{is}\:\mathrm{243}\left(=\mathrm{3}^{\mathrm{5}} \right) \\ $$$$\:\:\:\left[\:\:\mathrm{2}^{\mathrm{5}} =\mathrm{32}\:\:\mathrm{2}-{digit}\:{number}\:×\right. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{5}} =\mathrm{243}\:\:\mathrm{3}-{digit}\:{number}\checkmark \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{4}^{\mathrm{5}} =\mathrm{1024}\:\:\mathrm{4}-{digit}\:{number}×\:\:\:\right] \\ $$$${Possible}\:{candidate}\:{is}\:{only}\:\mathrm{243}\:{and} \\ $$$${it}\:{also}\:{fulfills}\:{the}\:{given}\:{condition}\:: \\ $$$$\:\:\:\left(\mathrm{243}\right)^{\mathrm{2}} \overset{?} {=}\left(\mathrm{2}+\mathrm{4}+\mathrm{3}\right)^{\mathrm{5}} \\ $$$$\:\:\:\:\:\left(\mathrm{3}^{\mathrm{5}} \right)^{\mathrm{2}} \overset{?} {=}\mathrm{9}^{\mathrm{5}} \\ $$$$\:\:\:\:\left(\mathrm{3}^{\mathrm{5}} \right)^{\mathrm{2}} \:\overset{?} {=}\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\mathrm{3}^{\mathrm{10}} =\mathrm{3}^{\mathrm{10}} \\ $$$$\therefore\:\mathrm{243}\:{is}\:{only}\:{successful}\:{candidate}. \\ $$$$\therefore\:{x}=\mathrm{2},{y}=\mathrm{4},{z}=\mathrm{3} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mid\left({x}+{y}+{z}\right)+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\mid \\ $$$$=\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} −\mid\left(\mathrm{2}+\mathrm{4}+\mathrm{3}\right)+\left(\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \mid\right. \\ $$$$=\mathrm{8}+\mathrm{64}+\mathrm{27}−\mid\mathrm{9}+\mathrm{29}\mid \\ $$$$=\mathrm{61} \\ $$
Commented by mathdanisur last updated on 19/Jul/21
$${cool}\:{Ser}\:{thanks} \\ $$