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Given-x-4-x-2-y-2-y-4-133-and-x-2-xy-y-2-7-then-what-is-the-value-of-xy-




Question Number 109577 by bemath last updated on 24/Aug/20
   Given x^4 +x^2 y^2 +y^4 =133                and x^2 −xy+y^2 =7    then what is the value of xy ?
$$\:\:\:\mathrm{G}{iven}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} =\mathrm{133} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\:\:{then}\:{what}\:{is}\:{the}\:{value}\:{of}\:{xy}\:? \\ $$
Commented by bemath last updated on 24/Aug/20
cooll.....
$${cooll}….. \\ $$
Answered by Dwaipayan Shikari last updated on 24/Aug/20
(x^2 +y^2 )^2 −x^2 y^2 =133  (7+xy)^2 −x^2 y^2 =133  49+14xy+x^2 y^2 −x^2 y^2 =133  14xy=84  xy=6
$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{133} \\ $$$$\left(\mathrm{7}+{xy}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{133} \\ $$$$\mathrm{49}+\mathrm{14}{xy}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{133} \\ $$$$\mathrm{14}{xy}=\mathrm{84} \\ $$$${xy}=\mathrm{6} \\ $$
Answered by john santu last updated on 24/Aug/20
    ((⇋JS⇌)/♠)  (1)x^4 +2x^2 y^2 +y^4 −x^2 y^2 =133  ⇒(x^2 +y^2 )^2 −(xy)^2 =133  ⇒(x^2 −xy+y^2 )(x^2 +xy+y^2 )=133  (2)x^2 −xy+y^2 =7  ⇒7(x^2 −xy+y^2 +2xy)=133  ⇒7(7+2xy)=133   ⇒7+2xy = 19 ⇒2xy=12    ∴ xy = 6
$$\:\:\:\:\frac{\leftrightharpoons{JS}\rightleftharpoons}{\spadesuit} \\ $$$$\left(\mathrm{1}\right){x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{133} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({xy}\right)^{\mathrm{2}} =\mathrm{133} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)=\mathrm{133} \\ $$$$\left(\mathrm{2}\right){x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\Rightarrow\mathrm{7}\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} +\mathrm{2}{xy}\right)=\mathrm{133} \\ $$$$\Rightarrow\mathrm{7}\left(\mathrm{7}+\mathrm{2}{xy}\right)=\mathrm{133}\: \\ $$$$\Rightarrow\mathrm{7}+\mathrm{2}{xy}\:=\:\mathrm{19}\:\Rightarrow\mathrm{2}{xy}=\mathrm{12} \\ $$$$\:\:\therefore\:{xy}\:=\:\mathrm{6} \\ $$
Commented by mnjuly1970 last updated on 24/Aug/20
Answered by Rasheed.Sindhi last updated on 24/Aug/20
((  (((x^2 −y^2 )(x^4 +x^2 y^2 +y^4 ))/(x^2 −y^2 ))  )/(((x+y)(x^2 −xy+y^2 ))/(x+y)))=((133)/7)=19  ((x^6 −y^6 )/(x^2 −y^2 ))×((x+y)/(x^3 +y^3 ))=19  (((x^3 −y^3 )(x^3 +y^3 ))/((x−y)(x^3 +y^3 )))=19  (((x−y)(x^2 +xy+y^2 ))/((x−y)))=19  x^2 +xy+y^2 =19........(i)   x^2 −xy+y^2 =7.........(ii)  (i)−(ii): 2xy=12                          xy=6
$$\frac{\:\:\frac{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} \right)}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }\:\:}{\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)}{{x}+{y}}}=\frac{\mathrm{133}}{\mathrm{7}}=\mathrm{19} \\ $$$$\frac{{x}^{\mathrm{6}} −{y}^{\mathrm{6}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }×\frac{{x}+{y}}{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }=\mathrm{19} \\ $$$$\frac{\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)}{\left({x}−{y}\right)\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)}=\mathrm{19} \\ $$$$\frac{\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)}{\left({x}−{y}\right)}=\mathrm{19} \\ $$$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{19}……..\left({i}\right) \\ $$$$\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7}………\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right):\:\mathrm{2}{xy}=\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xy}=\mathrm{6} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 24/Aug/20
   x^4 +x^2 y^2 +y^4 =133.........(i)     x^2 −xy+y^2 =7..............(ii)  ((x^4 +x^2 y^2 +y^4 )/( x^2 −xy+y^2 ))=((133)/7)=19  (((x^2 −xy+y^2 )(x^2 +xy+y^2 ))/(x^2 −xy+y^2 ))=19  x^2 +xy+y^2 =19..............(iii)  (iii)−(ii):2xy=12                           xy=6
$$\:\:\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} =\mathrm{133}………\left({i}\right) \\ $$$$\:\:\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7}…………..\left({ii}\right) \\ $$$$\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} }{\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} }=\frac{\mathrm{133}}{\mathrm{7}}=\mathrm{19} \\ $$$$\frac{\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} }=\mathrm{19} \\ $$$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{19}…………..\left({iii}\right) \\ $$$$\left({iii}\right)−\left({ii}\right):\mathrm{2}{xy}=\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xy}=\mathrm{6} \\ $$

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