Question Number 171944 by infinityaction last updated on 22/Jun/22
$${if} \\ $$$${a}+{b}+{c}=\mathrm{2196} \\ $$$$\sqrt[{\mathrm{3}}]{{a}}\:+{b}+{c}=\mathrm{2076} \\ $$$${a}+\sqrt[{\mathrm{3}}]{{b}}\:+{c}=\mathrm{1860} \\ $$$${a}+{b}+\sqrt[{\mathrm{3}}]{{c}}\:=\mathrm{480},\:{determine}\:{the}\:{value}\:{of} \\ $$$${a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} +{c}^{\frac{\mathrm{2}}{\mathrm{3}}} ,\:{if}\:{a},{b},{c}\:{are}\:{all}\:{integer}. \\ $$
Commented by infinityaction last updated on 22/Jun/22
$${repost}\:{of}\:{que}.\:{no}.\:\mathrm{171919} \\ $$
Commented by infinityaction last updated on 22/Jun/22
$$\mathrm{218} \\ $$
Commented by infinityaction last updated on 22/Jun/22
$${let}\:\:^{\mathrm{3}} \sqrt{{a}}={x},^{\mathrm{3}} \sqrt{{b}}\:=\:{y}\:{and}\:\:\:^{\mathrm{3}} \sqrt{{c}}\:\:=\:{z} \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{2196}\:\:\:\:\:\:\:\:…….\left(\mathrm{1}\right) \\ $$$$\:\:\:\:{x}+{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{2076}\:\:\:\:\:\:……….\left(\mathrm{2}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +{y}\:+{z}^{\mathrm{3}} =\mathrm{1860}\:\:\:\:\:\:………\left(\mathrm{3}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}=\mathrm{480}\:\:\:\:\:\:\:\:\:…….\left(\mathrm{4}\right) \\ $$$$\:\:\:\:{eq}^{{n}} \left(\mathrm{1}\right)−{eq}^{{n}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{3}} −{x}\:=\:\mathrm{120} \\ $$$$\:\:\:\left({x}−\mathrm{1}\right){x}\left({x}+\mathrm{1}\right)\:\:=\:\:\mathrm{120} \\ $$$$\:\:\:\:{a},{b},{c}\:\in\:{Z}\:\:\:{then}\:{x},{y},{z}\:\in\:{Z} \\ $$$$\:\:\:\left({x}−\mathrm{1}\right),{x},\left({x}+\mathrm{1}\right)\:\:{are}\:{in}\:{A}.{P} \\ $$$$\:\:{so}\:{factor}\:{of}\:\mathrm{120}\:=\:\mathrm{4}×\mathrm{5}×\mathrm{6} \\ $$$$\:\:{so}\:{x}\:=\:\mathrm{5} \\ $$$$\:\:{similarly} \\ $$$$\:\:\left({y}−\mathrm{1}\right){y}\left({y}+\mathrm{1}\right)\:=\:\mathrm{6}×\mathrm{7}×\mathrm{8}\:=\:\mathrm{336} \\ $$$$\:\:\:\:\:\:\:\:{y}\:\:\:\:=\:\:\:\mathrm{7} \\ $$$$\:\:\:{and} \\ $$$$\:\:\left({z}−\mathrm{1}\right){z}\left({z}+\mathrm{1}\right)\:=\:\mathrm{11}×\mathrm{12}×\mathrm{13}=\:\mathrm{1716} \\ $$$$\:\:\:\:\:\:\:{z}\:\:=\:\:\mathrm{12} \\ $$$$\:\:\:\:^{\mathrm{3}} \sqrt{{a}}\:\:=\:{x}\:\:\Rightarrow\:{a}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:{x}^{\mathrm{2}} \\ $$$$\:\:\:{a}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{25}\:\:\:….\left(\mathrm{4}\right) \\ $$$$\:\:\:\:\:{similarly} \\ $$$$\:\:\:\:\:\:\:{b}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{49}\:\:…..\left(\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:{and}\:\:\:{c}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{144}\:\:…\left(\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:{eq}^{{n}} \left(\mathrm{4}\right)+{eq}^{{n}} \left(\mathrm{5}\right)+{eq}^{{n}} \left(\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} +{c}^{\mathrm{2}/\mathrm{3}} \:=\:\mathrm{25}+\mathrm{49}+\mathrm{144} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} +{c}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{218} \\ $$
Commented by Tawa11 last updated on 22/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${thanks}\:{sir} \\ $$