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Question Number 61744 by Cheyboy last updated on 08/Jun/19

3xy^2 +x^3 =9 −−−−−(1)  3x^2 y+y^3 =18−−−−(2)  Find x and y

$$\mathrm{3}{xy}^{\mathrm{2}} +{x}^{\mathrm{3}} =\mathrm{9}\:−−−−−\left(\mathrm{1}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} {y}+{y}^{\mathrm{3}} =\mathrm{18}−−−−\left(\mathrm{2}\right) \\ $$$${Find}\:{x}\:{and}\:{y} \\ $$

Commented by Cheyboy last updated on 08/Jun/19

thank Godbless you

$${thank}\:{Godbless}\:{you} \\ $$

Commented by Cheyboy last updated on 08/Jun/19

sir how did the 3 disapper

$${sir}\:{how}\:{did}\:{the}\:\mathrm{3}\:{disapper} \\ $$

Commented by Prithwish sen last updated on 08/Jun/19

Adding (1) and (2)  (x+y)^3 =27  ∴ x+y =3.......(3)  Substracting (2) from (1)  (x−y)^3 =−9  ∴ x−y =−9^(1/3) .......(4)  Adding (3) and (4)  x = (1/2)(3−9^(1/3) )  Substracting (4)from (3)  y = (1/2)(3+9^(1/3) )

$$\mathrm{Adding}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} =\mathrm{27} \\ $$$$\therefore\:\mathrm{x}+\mathrm{y}\:=\mathrm{3}.......\left(\mathrm{3}\right) \\ $$$$\mathrm{Substracting}\:\left(\mathrm{2}\right)\:\mathrm{from}\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{3}} =−\mathrm{9} \\ $$$$\therefore\:\mathrm{x}−\mathrm{y}\:=−\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{3}}} .......\left(\mathrm{4}\right) \\ $$$$\mathrm{Adding}\:\left(\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4}\right) \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}−\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$\mathrm{Substracting}\:\left(\mathrm{4}\right)\mathrm{from}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\mathrm{9}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$ \\ $$

Commented by gunawan last updated on 08/Jun/19

nice solution

$$\mathrm{nice}\:\mathrm{solution} \\ $$

Commented by Prithwish sen last updated on 08/Jun/19

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Rasheed.Sindhi last updated on 08/Jun/19

e^x cellent!

$$\mathfrak{e}^{\mathrm{x}} \mathfrak{cellent}! \\ $$

Commented by Prithwish sen last updated on 08/Jun/19

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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