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Question Number 171728 by Mikenice last updated on 20/Jun/22

((x^3 +y^3 )/(x+y))=7 ((x^3 −y^3 )/(x−y))=19. find x and y

$$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}=\mathrm{7} \\ $$$$\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}−{y}}=\mathrm{19}.\:{find}\:{x}\:{and}\:{y} \\ $$

Answered by Ar Brandon last updated on 20/Jun/22

((x^3 +y^3 )/(x+y))=(((x+y)(x^2 −xy+y^2 ))/(x+y))=x^2 −xy+y^2 =7 ...eqn(i) ((x^3 −y^3 )/(x−y))=(((x−y)(x^2 +xy+y^2 ))/(x−y))=x^2 +xy+y^2 =19 ...eqn(ii) eqn(ii)− eqn(i) (x^2 +xy+y^2 )−(x^2 −xy+y^2 )=19−7 ⇒2xy = 12 ⇒xy=6 (At this point we can deduce (3, 2), (2, 3) are the solutions. But we can also get the result by solving.) ⇒y=(6/x) ...eqn(iii) Replacing eqn(iii) in eqn(i):- x^2 −x((6/x))+((6/x))^2 =7 ⇒x^2 −6+((36)/x^2 )=7 ⇒x^4 −13x^2 +36=0, (x^2 −9)(x^2 −4)=0 ⇒(x−3)(x+3)(x−2)(x+2)=0 ⇒x=3; x=−3; x=2; x=−2. Replacing the values of x in eqn(iii) to get corresponding values of y ⇒y=2; y=−2; y=3; y=−3. S={(x, y)∈R∣(3, 2); (−3, −2); (2, 3); (−2, −3)}

$$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}=\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)}{{x}+{y}}={x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7}\:...\mathrm{eqn}\left({i}\right) \\ $$$$\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}−{y}}=\frac{\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)}{{x}−{y}}={x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{19}\:...\mathrm{eqn}\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{eqn}\left({ii}\right)−\:\mathrm{eqn}\left({i}\right) \\ $$$$\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)−\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)=\mathrm{19}−\mathrm{7} \\ $$$$\Rightarrow\mathrm{2}{xy}\:=\:\mathrm{12}\:\Rightarrow{xy}=\mathrm{6}\:\left(\mathrm{At}\:\mathrm{this}\:\mathrm{point}\:\mathrm{we}\:\mathrm{can}\:\mathrm{deduce}\:\left(\mathrm{3},\:\mathrm{2}\right),\:\right. \\ $$$$\left.\left(\mathrm{2},\:\mathrm{3}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{solutions}.\:\mathrm{But}\:\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}\:\mathrm{by}\:\mathrm{solving}.\right) \\ $$$$\:\Rightarrow{y}=\frac{\mathrm{6}}{{x}}\:...\mathrm{eqn}\left({iii}\right) \\ $$$$\:\:\mathrm{Replacing}\:\mathrm{eqn}\left({iii}\right)\:\mathrm{in}\:\mathrm{eqn}\left({i}\right):- \\ $$$${x}^{\mathrm{2}} −{x}\left(\frac{\mathrm{6}}{{x}}\right)+\left(\frac{\mathrm{6}}{{x}}\right)^{\mathrm{2}} =\mathrm{7}\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{6}+\frac{\mathrm{36}}{{x}^{\mathrm{2}} }=\mathrm{7} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −\mathrm{13}{x}^{\mathrm{2}} +\mathrm{36}=\mathrm{0},\:\left({x}^{\mathrm{2}} −\mathrm{9}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3};\:{x}=−\mathrm{3};\:{x}=\mathrm{2};\:{x}=−\mathrm{2}. \\ $$$$\mathrm{Replacing}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{in}\:\mathrm{eqn}\left({iii}\right)\:\mathrm{to}\:\mathrm{get}\:\mathrm{corresponding}\: \\ $$$$\mathrm{values}\:\mathrm{of}\:{y} \\ $$$$\Rightarrow{y}=\mathrm{2};\:{y}=−\mathrm{2};\:{y}=\mathrm{3};\:{y}=−\mathrm{3}. \\ $$$$\mathrm{S}=\left\{\left({x},\:{y}\right)\in\mathbb{R}\mid\left(\mathrm{3},\:\mathrm{2}\right);\:\left(−\mathrm{3},\:−\mathrm{2}\right);\:\left(\mathrm{2},\:\mathrm{3}\right);\:\left(−\mathrm{2},\:−\mathrm{3}\right)\right\} \\ $$

Commented by Mikenice last updated on 20/Jun/22

thanks sir

$${thanks}\:{sir} \\ $$

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