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solve-in-R-x-3-y-3-19-xy-6-




Question Number 153698 by mathocean1 last updated on 09/Sep/21
solve in R   { ((x^3 −y^3 =19)),((xy=6)) :}
$${solve}\:{in}\:\mathbb{R} \\ $$$$\begin{cases}{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} =\mathrm{19}}\\{{xy}=\mathrm{6}}\end{cases} \\ $$
Answered by amin96 last updated on 09/Sep/21
(x−y)(x^2 +xy+y^2 )=19 ⇒ (x−y)((x−y)^2 +3xy)=19  (x−y)^3 +18(x−y)−19=0  ⇒  x−y=1  (x−y)((x+y)^2 −xy)=19   ⇒  (x+y)^2 =25   { ((x+y=±5)),((x−y=1 )) :}   ⇒   { ((x+y=5)),((x−y=1)) :}  ⇒  { ((x=3)),((y=2)) :}   { ((x+y=−5)),((x−y=1)) :}  ⇒  { ((x=−2)),((y=−3)) :}  ⇒ answer  (3; 2) (−2; −3)
$$\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)=\mathrm{19}\:\Rightarrow\:\left({x}−{y}\right)\left(\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{3}{xy}\right)=\mathrm{19} \\ $$$$\left({x}−{y}\right)^{\mathrm{3}} +\mathrm{18}\left({x}−{y}\right)−\mathrm{19}=\mathrm{0}\:\:\Rightarrow\:\:{x}−{y}=\mathrm{1} \\ $$$$\left({x}−{y}\right)\left(\left({x}+{y}\right)^{\mathrm{2}} −{xy}\right)=\mathrm{19}\:\:\:\Rightarrow\:\:\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$\begin{cases}{{x}+{y}=\pm\mathrm{5}}\\{{x}−{y}=\mathrm{1}\:}\end{cases}\:\:\:\Rightarrow\:\:\begin{cases}{{x}+{y}=\mathrm{5}}\\{{x}−{y}=\mathrm{1}}\end{cases}\:\:\Rightarrow\:\begin{cases}{{x}=\mathrm{3}}\\{{y}=\mathrm{2}}\end{cases} \\ $$$$\begin{cases}{{x}+{y}=−\mathrm{5}}\\{{x}−{y}=\mathrm{1}}\end{cases}\:\:\Rightarrow\:\begin{cases}{{x}=−\mathrm{2}}\\{{y}=−\mathrm{3}}\end{cases}\:\:\Rightarrow\:{answer}\:\:\left(\mathrm{3};\:\mathrm{2}\right)\:\left(−\mathrm{2};\:−\mathrm{3}\right) \\ $$
Answered by mr W last updated on 09/Sep/21
x^3 −((6/x))^3 =19  x^6 −19x^3 −216=0  x^3 =−8 or 27  ⇒x=−2 or 3  ⇒y=−3 or 2
$${x}^{\mathrm{3}} −\left(\frac{\mathrm{6}}{{x}}\right)^{\mathrm{3}} =\mathrm{19} \\ $$$${x}^{\mathrm{6}} −\mathrm{19}{x}^{\mathrm{3}} −\mathrm{216}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} =−\mathrm{8}\:{or}\:\mathrm{27} \\ $$$$\Rightarrow{x}=−\mathrm{2}\:{or}\:\mathrm{3} \\ $$$$\Rightarrow{y}=−\mathrm{3}\:{or}\:\mathrm{2} \\ $$
Commented by mathocean1 last updated on 09/Sep/21
thanks sirs
$${thanks}\:{sirs} \\ $$

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