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3xy-2x-y-3bx-3c-0-3xy-x-2y-3by-3c-0-find-non-zero-real-values-of-x-y-if-b-c-R-




Question Number 79613 by ajfour last updated on 26/Jan/20
3xy(2x−y)−3bx+3c=0  3xy(x−2y)−3by−3c=0  find non-zero, real values  of x,y  if b,c∈R.
$$\mathrm{3}{xy}\left(\mathrm{2}{x}−{y}\right)−\mathrm{3}{bx}+\mathrm{3}{c}=\mathrm{0} \\ $$$$\mathrm{3}{xy}\left({x}−\mathrm{2}{y}\right)−\mathrm{3}{by}−\mathrm{3}{c}=\mathrm{0} \\ $$$${find}\:{non}-{zero},\:{real}\:{values} \\ $$$${of}\:{x},{y}\:\:{if}\:{b},{c}\in\mathbb{R}. \\ $$
Answered by behi83417@gmail.com last updated on 26/Jan/20
xy(2x−y)=bx−c  xy(x−2y)=by+c  ⇒ { ((b(x+y)=3xy(x−y))),((xy(x+y)+2c=b(x−y))) :}  ⇒ { ((1.x−y=0⇒x=y⇒x^3 −bx−c=0)),((2.((b(x+y))/(xy(x+y)+2c))=((3xy)/b))) :}  [x^3 −bx−c=0,  △=B^2 −3AC=3b  k=((27c)/(2(√(27b^3 ))))=((9c)/(2b(√(3b))))  1.if :b>0,∣k∣≤1⇒ { ((x_1 =2(√(b/3))cos(((cos^(−1) (((9c)/(2b(√(3b))))))/3)))),((x_(2,3) =2(√(b/3))cos(((cos^(−1) (((9c)/(2b(√(3b)))))±((2π)/3))/3)))) :}  2.if:b>0,k>1⇒x=y=(√(b/3)).[((((9∣c∣)/(2b(√(3b))))+(√(((27c^2 )/(4b^3 ))−1))))^(1/3) +((((9∣c∣)/(2b(√(3b))))−(√(((27c^2 )/(4b^3 ))−1))))^(1/3) ]  3.if:b<0⇒x=y=(√((∣b∣)/3)).[((((9c)/(2b(√(3∣b∣))))+(√(((27c^2 )/(4∣b^3 ∣))+1))))^(1/3) +((((9c)/(2b(√(3∣b∣))))−(√(((27c^2 )/(4∣b^3 ∣))+1))))^(1/3) ]  4.b=0⇒x=y=(c)^(1/3)   −−−−−−−−−−+−−−−−−−  ((b(x+y))/(xy(x+y)+2c))=((3xy)/b)⇒  b^2 (x+y)=3(xy)^2 (x+y)+6cxy  ⇒3(x+y)(xy)^2 +6cxy−b(x+y)=0  ⇒(x+y)[b−3x^2 y^2 ]=6cxy⇒^(xy=p) x+y=((6cp)/(b−3p^2 ))  ⇒b^2 (((6cp)/(b−3p^2 )))^2 =9p^2 [(((6cp)/(b−3p^2 )))^2 −4p]  ⇒36c^2 p^2 b^2 =9p^2 [36c^2 p^2 −4p(b−3p^2 )^2 ]  [⇒p=0⇒no answer for x and y]  ⇒c^2 b^2 =9c^2 p^2 −pb^2 +6bp^2 −p^5   ⇒p^5 −3(2b+3c^2 )p^2 +b^2 p+c^2 b^2 =0  i can′t solve this......
$$\mathrm{xy}\left(\mathrm{2x}−\mathrm{y}\right)=\mathrm{bx}−\mathrm{c} \\ $$$$\mathrm{xy}\left(\mathrm{x}−\mathrm{2y}\right)=\mathrm{by}+\mathrm{c} \\ $$$$\Rightarrow\begin{cases}{\mathrm{b}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{3xy}\left(\mathrm{x}−\mathrm{y}\right)}\\{\mathrm{xy}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{2c}=\mathrm{b}\left(\mathrm{x}−\mathrm{y}\right)}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{1}.\mathrm{x}−\mathrm{y}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{y}\Rightarrow\mathrm{x}^{\mathrm{3}} −\mathrm{bx}−\mathrm{c}=\mathrm{0}}\\{\mathrm{2}.\frac{\mathrm{b}\left(\mathrm{x}+\mathrm{y}\right)}{\mathrm{xy}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{2c}}=\frac{\mathrm{3xy}}{\mathrm{b}}}\end{cases} \\ $$$$\left[\mathrm{x}^{\mathrm{3}} −\mathrm{bx}−\mathrm{c}=\mathrm{0},\:\:\bigtriangleup=\mathrm{B}^{\mathrm{2}} −\mathrm{3AC}=\mathrm{3b}\right. \\ $$$$\mathrm{k}=\frac{\mathrm{27c}}{\mathrm{2}\sqrt{\mathrm{27b}^{\mathrm{3}} }}=\frac{\mathrm{9c}}{\mathrm{2b}\sqrt{\mathrm{3b}}} \\ $$$$\mathrm{1}.\mathrm{if}\::\mathrm{b}>\mathrm{0},\mid\mathrm{k}\mid\leqslant\mathrm{1}\Rightarrow\begin{cases}{\mathrm{x}_{\mathrm{1}} =\mathrm{2}\sqrt{\frac{\mathrm{b}}{\mathrm{3}}}\mathrm{cos}\left(\frac{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{9c}}{\mathrm{2b}\sqrt{\mathrm{3b}}}\right)}{\mathrm{3}}\right)}\\{\mathrm{x}_{\mathrm{2},\mathrm{3}} =\mathrm{2}\sqrt{\frac{\mathrm{b}}{\mathrm{3}}}\mathrm{cos}\left(\frac{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{9c}}{\mathrm{2b}\sqrt{\mathrm{3b}}}\right)\pm\frac{\mathrm{2}\pi}{\mathrm{3}}}{\mathrm{3}}\right)}\end{cases} \\ $$$$\mathrm{2}.\mathrm{if}:\mathrm{b}>\mathrm{0},\mathrm{k}>\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{y}=\sqrt{\frac{\mathrm{b}}{\mathrm{3}}}.\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}\mid\mathrm{c}\mid}{\mathrm{2b}\sqrt{\mathrm{3b}}}+\sqrt{\frac{\mathrm{27c}^{\mathrm{2}} }{\mathrm{4b}^{\mathrm{3}} }−\mathrm{1}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}\mid\mathrm{c}\mid}{\mathrm{2b}\sqrt{\mathrm{3b}}}−\sqrt{\frac{\mathrm{27c}^{\mathrm{2}} }{\mathrm{4b}^{\mathrm{3}} }−\mathrm{1}}}\right] \\ $$$$\mathrm{3}.\mathrm{if}:\mathrm{b}<\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{y}=\sqrt{\frac{\mid\mathrm{b}\mid}{\mathrm{3}}}.\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{9c}}{\mathrm{2b}\sqrt{\mathrm{3}\mid\mathrm{b}\mid}}+\sqrt{\frac{\mathrm{27c}^{\mathrm{2}} }{\mathrm{4}\mid\mathrm{b}^{\mathrm{3}} \mid}+\mathrm{1}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9c}}{\mathrm{2b}\sqrt{\mathrm{3}\mid\mathrm{b}\mid}}−\sqrt{\frac{\mathrm{27c}^{\mathrm{2}} }{\mathrm{4}\mid\mathrm{b}^{\mathrm{3}} \mid}+\mathrm{1}}}\right] \\ $$$$\mathrm{4}.\mathrm{b}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{y}=\sqrt[{\mathrm{3}}]{\mathrm{c}} \\ $$$$−−−−−−−−−−+−−−−−−− \\ $$$$\frac{\mathrm{b}\left(\mathrm{x}+\mathrm{y}\right)}{\mathrm{xy}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{2c}}=\frac{\mathrm{3xy}}{\mathrm{b}}\Rightarrow \\ $$$$\mathrm{b}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{y}\right)=\mathrm{3}\left(\mathrm{xy}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{y}\right)+\mathrm{6cxy} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{xy}\right)^{\mathrm{2}} +\mathrm{6cxy}−\mathrm{b}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)\left[\mathrm{b}−\mathrm{3x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \right]=\mathrm{6cxy}\overset{\mathrm{xy}=\mathrm{p}} {\Rightarrow}\mathrm{x}+\mathrm{y}=\frac{\mathrm{6cp}}{\mathrm{b}−\mathrm{3p}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{b}^{\mathrm{2}} \left(\frac{\mathrm{6cp}}{\mathrm{b}−\mathrm{3p}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{9p}^{\mathrm{2}} \left[\left(\frac{\mathrm{6cp}}{\mathrm{b}−\mathrm{3p}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{4p}\right] \\ $$$$\Rightarrow\mathrm{36c}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} =\mathrm{9p}^{\mathrm{2}} \left[\mathrm{36c}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} −\mathrm{4p}\left(\mathrm{b}−\mathrm{3p}^{\mathrm{2}} \right)^{\mathrm{2}} \right] \\ $$$$\left[\Rightarrow\mathrm{p}=\mathrm{0}\Rightarrow\mathrm{no}\:\mathrm{answer}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\right] \\ $$$$\Rightarrow\mathrm{c}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} =\mathrm{9c}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} −\mathrm{pb}^{\mathrm{2}} +\mathrm{6bp}^{\mathrm{2}} −\mathrm{p}^{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{5}} −\mathrm{3}\left(\mathrm{2b}+\mathrm{3c}^{\mathrm{2}} \right)\mathrm{p}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \mathrm{p}+\mathrm{c}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{this}…… \\ $$
Commented by ajfour last updated on 26/Jan/20
Thanks Sir, for such a general  solution..
$${Thanks}\:{Sir},\:{for}\:{such}\:{a}\:{general} \\ $$$${solution}.. \\ $$

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