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If-x-3y-6z-7z-2-and-y-1-2-z-6y-3-2-z-6y-Find-x-3-y-3-

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Question Number 9009 by tawakalitu last updated on 13/Nov/16
If : x = ((3y + 6z)/(7z − 2)) and y = (((1/2)z + 6y)/((3/2)z + 6y)) Find : x^3 + y^3
$$\mathrm{If}\::\:\:\mathrm{x}\:=\:\frac{\mathrm{3y}\:+\:\mathrm{6z}}{\mathrm{7z}\:−\:\mathrm{2}}\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}\:+\:\mathrm{6y}}{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{z}\:+\:\mathrm{6y}} \\ $$$$\mathrm{Find}\::\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \\ $$
Answered by Rasheed Soomro last updated on 14/Nov/16
If : x = ((3y + 6z)/(7z − 2)) and y = (((1/2)z + 6y)/((3/2)z + 6y))=((z+12y)/(3z+12y)) y=((z+12y)/(3z+12y))⇒3yz+12y^2 =z+12y 12y^2 +3(z−4)y−z=0 y=((−3(z−4)±(√(9(z−4)^2 +48z)))/(24)) =((−3(z−4)±(√(9z^2 −24z+144)))/(24)) −−−− x = ((3y + 6z)/(7z − 2)) =((3(((−3(z−4)±(√(9z^2 −24z+144)))/(24)))+6z)/(7z−2)) =((((−3(z−4)±(√(9z^2 −24z+144)))/8)+6z)/(7z−2)) =(((−3(z−4)±(√(9z^2 −24z+144))+48z)/8)/(7z−2)) =((45z+12±(√(9z^2 −24z+144)))/(8(7z−2))) x^3 +y^3 =(((45z+12±(√(9z^2 −24z+144)))/(8(7z−2))))^3 +(((−3(z−4)±(√(9z^2 −24z+144)))/(24)))^3
$$\mathrm{If}\::\:\:\mathrm{x}\:=\:\frac{\mathrm{3y}\:+\:\mathrm{6z}}{\mathrm{7z}\:−\:\mathrm{2}}\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{z}\:+\:\mathrm{6y}}{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{z}\:+\:\mathrm{6y}}=\frac{\mathrm{z}+\mathrm{12y}}{\mathrm{3z}+\mathrm{12y}} \\ $$$$\mathrm{y}=\frac{\mathrm{z}+\mathrm{12y}}{\mathrm{3z}+\mathrm{12y}}\Rightarrow\mathrm{3yz}+\mathrm{12y}^{\mathrm{2}} =\mathrm{z}+\mathrm{12y} \\ $$$$\mathrm{12y}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\mathrm{y}−\mathrm{z}=\mathrm{0} \\ $$$$\mathrm{y}=\frac{−\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\pm\sqrt{\mathrm{9}\left(\mathrm{z}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{48z}}}{\mathrm{24}} \\ $$$$\:\:\:\:=\frac{−\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}}{\mathrm{24}} \\ $$$$−−−− \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{3y}\:+\:\mathrm{6z}}{\mathrm{7z}\:−\:\mathrm{2}}\:=\frac{\mathrm{3}\left(\frac{−\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}}{\mathrm{24}}\right)+\mathrm{6z}}{\mathrm{7z}−\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\frac{−\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}}{\mathrm{8}}+\mathrm{6z}}{\mathrm{7z}−\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\frac{−\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}+\mathrm{48z}}{\mathrm{8}}}{\mathrm{7z}−\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{45z}+\mathrm{12}\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}}{\mathrm{8}\left(\mathrm{7z}−\mathrm{2}\right)} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\left(\frac{\mathrm{45z}+\mathrm{12}\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}}{\mathrm{8}\left(\mathrm{7z}−\mathrm{2}\right)}\right)^{\mathrm{3}} +\left(\frac{−\mathrm{3}\left(\mathrm{z}−\mathrm{4}\right)\pm\sqrt{\mathrm{9z}^{\mathrm{2}} −\mathrm{24z}+\mathrm{144}}}{\mathrm{24}}\right)^{\mathrm{3}} \\ $$$$\: \\ $$
Commented by tawakalitu last updated on 13/Nov/16
thanks sir. will be expecting.
$$\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{will}\:\mathrm{be}\:\mathrm{expecting}. \\ $$
Commented by tawakalitu last updated on 14/Nov/16
i really appreciate sir. God bless you.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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