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Question Number 204421 by necx122 last updated on 17/Feb/24

x^3 +y^3 = 35 (1/x)+(1/y) =(5/6) solve for all possible values of x and y

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} \:=\:\mathrm{35} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${solve}\:{for}\:{all}\:{possible}\:{values}\:{of}\:{x}\:{and}\:{y} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 17/Feb/24

x^3 +y^3 = 35 ⇒(1/x^3 )+(1/y^3 )=((35)/(x^3 y^3 )) ⇒((1/x)+(1/y))^3 −3((1/x))((1/y))((1/x)+(1/y))=((35)/(x^3 y^3 )) ((5/6))^3 −(3/(xy))((5/6))=((35)/(x^3 y^3 )) ((125)/(216))−(5/(2xy))=((35)/(x^3 y^3 )) ((125xy−540)/(216xy))=((35)/(x^3 y^3 )) ((125xy−540)/(216))=((35)/(x^2 y^2 )) 125x^3 y^3 −540x^2 y^2 =35×216 25(xy)^3 −108(xy)^2 −1512=0 (xy−6)( 25(xy)^2 +42x+252 )=0 xy=6 ,((−42±(√(42^2 −4(25)(252))))/(50)) xy=6 ,((−21±3i(√(651)))/(25)) (1/x)+(1/y) =(5/6) ((x+y)/(xy))=(5/6) ((x+y)/6)=(5/6) x+y=5 xy=6 ∧ x+y=5⇒(x,y)=(2,3),(3,2)

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} \:=\:\mathrm{35} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{y}^{\mathrm{3}} }=\frac{\mathrm{35}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} } \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{1}}{{x}}\right)\left(\frac{\mathrm{1}}{{y}}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)=\frac{\mathrm{35}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} } \\ $$$$\:\:\:\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} −\frac{\mathrm{3}}{{xy}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)=\frac{\mathrm{35}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} } \\ $$$$\:\:\:\frac{\mathrm{125}}{\mathrm{216}}−\frac{\mathrm{5}}{\mathrm{2}{xy}}=\frac{\mathrm{35}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} } \\ $$$$\:\:\:\frac{\mathrm{125}{xy}−\mathrm{540}}{\mathrm{216}{xy}}=\frac{\mathrm{35}}{{x}^{\mathrm{3}} {y}^{\mathrm{3}} } \\ $$$$\:\:\:\frac{\mathrm{125}{xy}−\mathrm{540}}{\mathrm{216}}=\frac{\mathrm{35}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{125}{x}^{\mathrm{3}} {y}^{\mathrm{3}} −\mathrm{540}{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{35}×\mathrm{216} \\ $$$$\:\:\:\mathrm{25}\left({xy}\right)^{\mathrm{3}} −\mathrm{108}\left({xy}\right)^{\mathrm{2}} −\mathrm{1512}=\mathrm{0} \\ $$$$\:\:\:\left({xy}−\mathrm{6}\right)\left(\:\mathrm{25}\left({xy}\right)^{\mathrm{2}} +\mathrm{42}{x}+\mathrm{252}\:\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{xy}=\mathrm{6}\:,\frac{−\mathrm{42}\pm\sqrt{\mathrm{42}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{25}\right)\left(\mathrm{252}\right)}}{\mathrm{50}} \\ $$$$\:\:\:\:\:{xy}=\mathrm{6}\:,\frac{−\mathrm{21}\pm\mathrm{3}\boldsymbol{{i}}\sqrt{\mathrm{651}}}{\mathrm{25}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\frac{{x}+{y}}{{xy}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\frac{{x}+{y}}{\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${x}+{y}=\mathrm{5} \\ $$$${xy}=\mathrm{6}\:\wedge\:{x}+{y}=\mathrm{5}\Rightarrow\left({x},{y}\right)=\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{3},\mathrm{2}\right) \\ $$

Commented by necx122 last updated on 17/Feb/24

Wow! This is lovely and detailed. Thank you for always clearing my doubt.

$${Wow}!\:{This}\:{is}\:{lovely}\:{and}\:{detailed}.\:{Thank}\:{you} \\ $$$${for}\:{always}\:{clearing}\:{my}\:{doubt}. \\ $$

Answered by Rasheed.Sindhi last updated on 17/Feb/24

(1/x)+(1/y) =(5/6)⇒((x+y)/(xy))=(5/6)⇒xy=((6(x+y))/5) x^3 +y^3 = 35⇒(x+y)^3 −3xy(x+y)=35 ⇒(x+y)^3 −3(((6(x+y))/5))(x+y)=35 5(x+y)^3 −18(x+y)^2 −175=0 5t^3 −18t^2 −175=0; [t=x+y] (t−5)(5t^2 +7t+35)=0 t=5,((−7±i(√(651)))/(10)) x+y=5,((−7±i(√(651)))/(10)) xy=((6(x+y))/5)=((6(5))/5),((6(((−7±i(√(651)))/(10))))/5) xy=6,((3(−7±i(√(651)) ))/(25)) x+y=5 ∧ xy=6 ⇒(x,y)=(2,5),(5,2) x+y=((−7±i(√(651)))/(10)) ∧ xy=((3(−7±i(√(651)) ))/(25)) y=((−7±i(√(651)))/(10))−x ∧ x(((−7±i(√(651)))/(10))−x)=((3(−7±i(√(651)) ))/(25)) ((−7±i(√(651)))/(10))x−x^2 =((3(−7±i(√(651)) ))/(25)) x^2 −((−7±i(√(651)))/(10))x+((3(−7±i(√(651)) ))/(25))=0 Continue

$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:=\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow\frac{{x}+{y}}{{xy}}=\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow{xy}=\frac{\mathrm{6}\left({x}+{y}\right)}{\mathrm{5}} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} \:=\:\mathrm{35}\Rightarrow\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{35} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{6}\left({x}+{y}\right)}{\mathrm{5}}\right)\left({x}+{y}\right)=\mathrm{35} \\ $$$$\mathrm{5}\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{18}\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{175}=\mathrm{0} \\ $$$$\mathrm{5}{t}^{\mathrm{3}} −\mathrm{18}{t}^{\mathrm{2}} −\mathrm{175}=\mathrm{0};\:\left[{t}={x}+{y}\right] \\ $$$$\left({t}−\mathrm{5}\right)\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{7}{t}+\mathrm{35}\right)=\mathrm{0} \\ $$$${t}=\mathrm{5},\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}} \\ $$$${x}+{y}=\mathrm{5},\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}} \\ $$$${xy}=\frac{\mathrm{6}\left({x}+{y}\right)}{\mathrm{5}}=\frac{\mathrm{6}\left(\mathrm{5}\right)}{\mathrm{5}},\frac{\mathrm{6}\left(\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}}\right)}{\mathrm{5}} \\ $$$${xy}=\mathrm{6},\frac{\mathrm{3}\left(−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}\:\right)}{\mathrm{25}} \\ $$$${x}+{y}=\mathrm{5}\:\wedge\:{xy}=\mathrm{6}\:\Rightarrow\left({x},{y}\right)=\left(\mathrm{2},\mathrm{5}\right),\left(\mathrm{5},\mathrm{2}\right) \\ $$$$\: \\ $$$${x}+{y}=\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}}\:\wedge\:{xy}=\frac{\mathrm{3}\left(−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}\:\right)}{\mathrm{25}} \\ $$$${y}=\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}}−{x} \\ $$$$\:\:\:\wedge\:{x}\left(\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}}−{x}\right)=\frac{\mathrm{3}\left(−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}\:\right)}{\mathrm{25}} \\ $$$$\:\:\:\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}}{x}−{x}^{\mathrm{2}} =\frac{\mathrm{3}\left(−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}\:\right)}{\mathrm{25}} \\ $$$$\:\:\:{x}^{\mathrm{2}} −\frac{−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}}{\mathrm{10}}{x}+\frac{\mathrm{3}\left(−\mathrm{7}\pm{i}\sqrt{\mathrm{651}}\:\right)}{\mathrm{25}}=\mathrm{0} \\ $$$${Continue} \\ $$

Answered by Frix last updated on 17/Feb/24

x^3 +y^3 =35 (1/x)+(1/y)=(5/6) ⇔ 5xy=6(x+y) x=u−v∧y=u+v 2u^3 +6uv^2 =35 5u^2 −5v^2 =12u ⇒ ((35−2u^3 )/(6u))=((5u^2 −12u)/5) u^3 −((9u^2 )/5)−((35)/8)=0 u=(5/2) ⇒ v=(1/2) u=−(7/(20))±((√(651))/(20))i ⇒ v=((√(−133+10(√(6433))))/(20))∓((√(133+10(√(6433))))/(20))i ⇒ x, y

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{35} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{5}}{\mathrm{6}}\:\Leftrightarrow\:\mathrm{5}{xy}=\mathrm{6}\left({x}+{y}\right) \\ $$$${x}={u}−{v}\wedge{y}={u}+{v} \\ $$$$\mathrm{2}{u}^{\mathrm{3}} +\mathrm{6}{uv}^{\mathrm{2}} =\mathrm{35} \\ $$$$\mathrm{5}{u}^{\mathrm{2}} −\mathrm{5}{v}^{\mathrm{2}} =\mathrm{12}{u} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{35}−\mathrm{2}{u}^{\mathrm{3}} }{\mathrm{6}{u}}=\frac{\mathrm{5}{u}^{\mathrm{2}} −\mathrm{12}{u}}{\mathrm{5}} \\ $$$${u}^{\mathrm{3}} −\frac{\mathrm{9}{u}^{\mathrm{2}} }{\mathrm{5}}−\frac{\mathrm{35}}{\mathrm{8}}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow\:{v}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${u}=−\frac{\mathrm{7}}{\mathrm{20}}\pm\frac{\sqrt{\mathrm{651}}}{\mathrm{20}}\mathrm{i}\:\Rightarrow\:{v}=\frac{\sqrt{−\mathrm{133}+\mathrm{10}\sqrt{\mathrm{6433}}}}{\mathrm{20}}\mp\frac{\sqrt{\mathrm{133}+\mathrm{10}\sqrt{\mathrm{6433}}}}{\mathrm{20}}\mathrm{i} \\ $$$$\Rightarrow\:{x},\:{y} \\ $$

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