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Question Number 121449 by liberty last updated on 08/Nov/20

   { ((x+y=3)),((x^5 +y^5 =33)) :}

$$\:\:\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{3}}\\{\mathrm{x}^{\mathrm{5}} +\mathrm{y}^{\mathrm{5}} =\mathrm{33}}\end{cases} \\ $$

Commented by Dwaipayan Shikari last updated on 08/Nov/20

x=1or 2  y=2 or 1      (By inspection)

$${x}=\mathrm{1}{or}\:\mathrm{2} \\ $$$${y}=\mathrm{2}\:{or}\:\mathrm{1}\:\:\:\:\:\:\left({By}\:{inspection}\right) \\ $$

Answered by MJS_new last updated on 08/Nov/20

x+y=a  x^5 +y^5 =b  let x=u−v∧y=u+v  (1) 2u=a ⇒ u=(a/2)  insert in (2)  5av^4 +((5a^3 )/2)v^2 +(a^5 /(16))=b  v^4 +(a^2 /2)v^2 +((a^5 −16b)/(80a))=0  v=±(√(−(a^2 /4)±((√(5(a^5 +4b)))/(10(√a)))))  [4 solutions]  in this case  u=(3/2)∧(v=±(1/2)∨v=±((√(19))/2)i)

$${x}+{y}={a} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} ={b} \\ $$$$\mathrm{let}\:{x}={u}−{v}\wedge{y}={u}+{v} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{u}={a}\:\Rightarrow\:{u}=\frac{{a}}{\mathrm{2}} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{5}{av}^{\mathrm{4}} +\frac{\mathrm{5}{a}^{\mathrm{3}} }{\mathrm{2}}{v}^{\mathrm{2}} +\frac{{a}^{\mathrm{5}} }{\mathrm{16}}={b} \\ $$$${v}^{\mathrm{4}} +\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{v}^{\mathrm{2}} +\frac{{a}^{\mathrm{5}} −\mathrm{16}{b}}{\mathrm{80}{a}}=\mathrm{0} \\ $$$${v}=\pm\sqrt{−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\pm\frac{\sqrt{\mathrm{5}\left({a}^{\mathrm{5}} +\mathrm{4}{b}\right)}}{\mathrm{10}\sqrt{{a}}}}\:\:\left[\mathrm{4}\:\mathrm{solutions}\right] \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${u}=\frac{\mathrm{3}}{\mathrm{2}}\wedge\left({v}=\pm\frac{\mathrm{1}}{\mathrm{2}}\vee{v}=\pm\frac{\sqrt{\mathrm{19}}}{\mathrm{2}}\mathrm{i}\right) \\ $$

Commented by liberty last updated on 08/Nov/20

yes..

$$\mathrm{yes}.. \\ $$

Answered by bramlexs22 last updated on 08/Nov/20

let x+y=u=3 ∧xy=v  (x+y)^5 =x^5 +5x^4 y+10x^3 y^2 +10x^2 y^3 +5xy^4 +y^5   ⇔243=33+5xy(x^3 +y^3 )+10(xy)^2 (x+y)  ⇒210=5v((x+y)^3 −3xy(x+y))+30v^2   ⇒210=5v(27−9v)+30v^2   ⇒42=27v−9v^2 +6v^2   ⇒3v^2 −9v+14=0 → { ((v=2)),((v=7)) :}  case(1) v=2 ∧u=3  ⇒xy=2 ∧x+y=3 → { ((x=2∧y=1)),((x=1∧y=2)) :}  case(2) xy=7 ∧x+y=3  ⇒x(3−x)=7 , x^2 −3x+7=0  → { ((x=((3+i(√(19)))/2) →y=((3−i(√(19)))/2))),((x=((3−i(√(19)))/2)→y=((3+i(√(19)))/2))) :}

$$\mathrm{let}\:\mathrm{x}+\mathrm{y}=\mathrm{u}=\mathrm{3}\:\wedge\mathrm{xy}=\mathrm{v} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{5}} =\mathrm{x}^{\mathrm{5}} +\mathrm{5x}^{\mathrm{4}} \mathrm{y}+\mathrm{10x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} +\mathrm{10x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} +\mathrm{5xy}^{\mathrm{4}} +\mathrm{y}^{\mathrm{5}} \\ $$$$\Leftrightarrow\mathrm{243}=\mathrm{33}+\mathrm{5xy}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} \right)+\mathrm{10}\left(\mathrm{xy}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{210}=\mathrm{5v}\left(\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} −\mathrm{3xy}\left(\mathrm{x}+\mathrm{y}\right)\right)+\mathrm{30v}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{210}=\mathrm{5v}\left(\mathrm{27}−\mathrm{9v}\right)+\mathrm{30v}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{42}=\mathrm{27v}−\mathrm{9v}^{\mathrm{2}} +\mathrm{6v}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3v}^{\mathrm{2}} −\mathrm{9v}+\mathrm{14}=\mathrm{0}\:\rightarrow\begin{cases}{\mathrm{v}=\mathrm{2}}\\{\mathrm{v}=\mathrm{7}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{v}=\mathrm{2}\:\wedge\mathrm{u}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{xy}=\mathrm{2}\:\wedge\mathrm{x}+\mathrm{y}=\mathrm{3}\:\rightarrow\begin{cases}{\mathrm{x}=\mathrm{2}\wedge\mathrm{y}=\mathrm{1}}\\{\mathrm{x}=\mathrm{1}\wedge\mathrm{y}=\mathrm{2}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{xy}=\mathrm{7}\:\wedge\mathrm{x}+\mathrm{y}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{3}−\mathrm{x}\right)=\mathrm{7}\:,\:\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{7}=\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{3}+{i}\sqrt{\mathrm{19}}}{\mathrm{2}}\:\rightarrow\mathrm{y}=\frac{\mathrm{3}−{i}\sqrt{\mathrm{19}}}{\mathrm{2}}}\\{{x}=\frac{\mathrm{3}−{i}\sqrt{\mathrm{19}}}{\mathrm{2}}\rightarrow\mathrm{y}=\frac{\mathrm{3}+{i}\sqrt{\mathrm{19}}}{\mathrm{2}}}\end{cases} \\ $$

Commented by liberty last updated on 08/Nov/20

gave kudos

$$\mathrm{gave}\:\mathrm{kudos} \\ $$

Answered by mr W last updated on 08/Nov/20

(x+y)^5 =x^5 +5x^4 y+10x^3 y^2 +10x^2 y^3 +5xy^4 +y^5   3^5 =33+5xy(x^3 +2x^2 y+2xy^2 +y^3 )  3^5 =33+5xy[x^3 +3x^2 y+3xy^2 +y^3 −xy(x+y)]  3^5 =33+5xy[(x+y)^3 −xy(x+y)]  3^5 =33+5xy(3^3 −3xy)  14=xy(9−xy)  (xy)^2 −9(xy)+14=0  (xy−2)(xy−7)=0  ⇒xy=2 or 7  z^2 −3z+2=0 ⇒(x,y)=1,2  z^2 −3z+7=0 ⇒(x,y)=((3+i(√(19)))/2),((3−i(√(19)))/2)

$$\left({x}+{y}\right)^{\mathrm{5}} ={x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{4}} {y}+\mathrm{10}{x}^{\mathrm{3}} {y}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +\mathrm{5}{xy}^{\mathrm{4}} +{y}^{\mathrm{5}} \\ $$$$\mathrm{3}^{\mathrm{5}} =\mathrm{33}+\mathrm{5}{xy}\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} {y}+\mathrm{2}{xy}^{\mathrm{2}} +{y}^{\mathrm{3}} \right) \\ $$$$\mathrm{3}^{\mathrm{5}} =\mathrm{33}+\mathrm{5}{xy}\left[{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} {y}+\mathrm{3}{xy}^{\mathrm{2}} +{y}^{\mathrm{3}} −{xy}\left({x}+{y}\right)\right] \\ $$$$\mathrm{3}^{\mathrm{5}} =\mathrm{33}+\mathrm{5}{xy}\left[\left({x}+{y}\right)^{\mathrm{3}} −{xy}\left({x}+{y}\right)\right] \\ $$$$\mathrm{3}^{\mathrm{5}} =\mathrm{33}+\mathrm{5}{xy}\left(\mathrm{3}^{\mathrm{3}} −\mathrm{3}{xy}\right) \\ $$$$\mathrm{14}={xy}\left(\mathrm{9}−{xy}\right) \\ $$$$\left({xy}\right)^{\mathrm{2}} −\mathrm{9}\left({xy}\right)+\mathrm{14}=\mathrm{0} \\ $$$$\left({xy}−\mathrm{2}\right)\left({xy}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow{xy}=\mathrm{2}\:{or}\:\mathrm{7} \\ $$$${z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}=\mathrm{0}\:\Rightarrow\left({x},{y}\right)=\mathrm{1},\mathrm{2} \\ $$$${z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{7}=\mathrm{0}\:\Rightarrow\left({x},{y}\right)=\frac{\mathrm{3}+{i}\sqrt{\mathrm{19}}}{\mathrm{2}},\frac{\mathrm{3}−{i}\sqrt{\mathrm{19}}}{\mathrm{2}} \\ $$

Commented by liberty last updated on 08/Nov/20

yess...

$$\mathrm{yess}... \\ $$

Commented by harckinwunmy last updated on 08/Nov/20

ff

$$\mathrm{ff} \\ $$

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