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Question Number 171944 by infinityaction last updated on 22/Jun/22

if a+b+c=2196 (a)^(1/3) +b+c=2076 a+(b)^(1/3) +c=1860 a+b+(c)^(1/3) =480, determine the value of a^(2/3) +b^(2/3) +c^(2/3) , if a,b,c are all integer.

$${if} \\ $$$${a}+{b}+{c}=\mathrm{2196} \\ $$$$\sqrt[{\mathrm{3}}]{{a}}\:+{b}+{c}=\mathrm{2076} \\ $$$${a}+\sqrt[{\mathrm{3}}]{{b}}\:+{c}=\mathrm{1860} \\ $$$${a}+{b}+\sqrt[{\mathrm{3}}]{{c}}\:=\mathrm{480},\:{determine}\:{the}\:{value}\:{of} \\ $$$${a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} +{c}^{\frac{\mathrm{2}}{\mathrm{3}}} ,\:{if}\:{a},{b},{c}\:{are}\:{all}\:{integer}. \\ $$

Commented by infinityaction last updated on 22/Jun/22

repost of que. no. 171919

$${repost}\:{of}\:{que}.\:{no}.\:\mathrm{171919} \\ $$

Commented by infinityaction last updated on 22/Jun/22

218

$$\mathrm{218} \\ $$

Commented by infinityaction last updated on 22/Jun/22

let ^3 (√a)=x,^3 (√b) = y and ^3 (√c) = z x^3 +y^3 +z^3 =2196 .......(1) x+y^3 +z^3 =2076 ..........(2) x^3 +y +z^3 =1860 .........(3) x^3 +y^3 +z=480 .......(4) eq^n (1)−eq^n (2) x^3 −x = 120 (x−1)x(x+1) = 120 a,b,c ∈ Z then x,y,z ∈ Z (x−1),x,(x+1) are in A.P so factor of 120 = 4×5×6 so x = 5 similarly (y−1)y(y+1) = 6×7×8 = 336 y = 7 and (z−1)z(z+1) = 11×12×13= 1716 z = 12 ^3 (√a) = x ⇒ a^(2/3) = x^2 a^(2/3) = 25 ....(4) similarly b^(2/3) = 49 .....(5) and c^(2/3) = 144 ...(6) eq^n (4)+eq^n (5)+eq^n (6) a^(2/3) +b^(2/3) +c^(2/3) = 25+49+144 a^(2/3) +b^(2/3) +c^(2/3) = 218

$${let}\:\:^{\mathrm{3}} \sqrt{{a}}={x},^{\mathrm{3}} \sqrt{{b}}\:=\:{y}\:{and}\:\:\:^{\mathrm{3}} \sqrt{{c}}\:\:=\:{z} \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{2196}\:\:\:\:\:\:\:\:.......\left(\mathrm{1}\right) \\ $$$$\:\:\:\:{x}+{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{2076}\:\:\:\:\:\:..........\left(\mathrm{2}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +{y}\:+{z}^{\mathrm{3}} =\mathrm{1860}\:\:\:\:\:\:.........\left(\mathrm{3}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}=\mathrm{480}\:\:\:\:\:\:\:\:\:.......\left(\mathrm{4}\right) \\ $$$$\:\:\:\:{eq}^{{n}} \left(\mathrm{1}\right)−{eq}^{{n}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{3}} −{x}\:=\:\mathrm{120} \\ $$$$\:\:\:\left({x}−\mathrm{1}\right){x}\left({x}+\mathrm{1}\right)\:\:=\:\:\mathrm{120} \\ $$$$\:\:\:\:{a},{b},{c}\:\in\:{Z}\:\:\:{then}\:{x},{y},{z}\:\in\:{Z} \\ $$$$\:\:\:\left({x}−\mathrm{1}\right),{x},\left({x}+\mathrm{1}\right)\:\:{are}\:{in}\:{A}.{P} \\ $$$$\:\:{so}\:{factor}\:{of}\:\mathrm{120}\:=\:\mathrm{4}×\mathrm{5}×\mathrm{6} \\ $$$$\:\:{so}\:{x}\:=\:\mathrm{5} \\ $$$$\:\:{similarly} \\ $$$$\:\:\left({y}−\mathrm{1}\right){y}\left({y}+\mathrm{1}\right)\:=\:\mathrm{6}×\mathrm{7}×\mathrm{8}\:=\:\mathrm{336} \\ $$$$\:\:\:\:\:\:\:\:{y}\:\:\:\:=\:\:\:\mathrm{7} \\ $$$$\:\:\:{and} \\ $$$$\:\:\left({z}−\mathrm{1}\right){z}\left({z}+\mathrm{1}\right)\:=\:\mathrm{11}×\mathrm{12}×\mathrm{13}=\:\mathrm{1716} \\ $$$$\:\:\:\:\:\:\:{z}\:\:=\:\:\mathrm{12} \\ $$$$\:\:\:\:^{\mathrm{3}} \sqrt{{a}}\:\:=\:{x}\:\:\Rightarrow\:{a}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:{x}^{\mathrm{2}} \\ $$$$\:\:\:{a}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{25}\:\:\:....\left(\mathrm{4}\right) \\ $$$$\:\:\:\:\:{similarly} \\ $$$$\:\:\:\:\:\:\:{b}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{49}\:\:.....\left(\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:{and}\:\:\:{c}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{144}\:\:...\left(\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:{eq}^{{n}} \left(\mathrm{4}\right)+{eq}^{{n}} \left(\mathrm{5}\right)+{eq}^{{n}} \left(\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} +{c}^{\mathrm{2}/\mathrm{3}} \:=\:\mathrm{25}+\mathrm{49}+\mathrm{144} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} +{c}^{\mathrm{2}/\mathrm{3}} \:\:=\:\:\mathrm{218} \\ $$

Commented by Tawa11 last updated on 22/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Mikenice last updated on 23/Jun/22

thanks sir

$${thanks}\:{sir} \\ $$

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