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Question Number 52592 by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

if x+y+z=1  x^2 +y^2 +z^2 =2  x^3 +y^3 +z^3 =3  find x^4 +y^4 +z^4

$${if}\:{x}+{y}+{z}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{3} \\ $$$${find}\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} \\ $$

Answered by math1967 last updated on 10/Jan/19

x+y+z=1  ⇒2(xy+yz+zx)=1−2=−1  ∴(xy+yz+zx)=−(1/2)  x^2 y^2 +y^2 z^2 +z^2 x^2 +2xyz(x+y+z)=(1/4)  x^2 y^2 +y^2 z^2 +z^2 x^2 =(1/4)−(1/3)=−(1/(12)) ★  x^3 +y^3 +z^3 −3xyz+3xyz=3★  (x+y+z)(x^2 +y^2 +z^2 −xy−yz−zx)+3xyz=3  3xyz=3−(5/2)=(1/2)⇒xyz=(1/6)  (x^2 +y^2 +z^2 )^2 =4  x^4 +y^4 +z^4 +2(x^2 y^2 +y^2 z^2 +z^2 x^2 )=4  x^4 +y^4 +z^4 =4+(1/6)=4(1/6)ans

$${x}+{y}+{z}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\left({xy}+{yz}+{zx}\right)=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$$$\therefore\left({xy}+{yz}+{zx}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{xyz}\left({x}+{y}+{z}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\bigstar \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz}+\mathrm{3}{xyz}=\mathrm{3}\bigstar \\ $$$$\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}\right)+\mathrm{3}{xyz}=\mathrm{3} \\ $$$$\mathrm{3}{xyz}=\mathrm{3}−\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{xyz}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} +\mathrm{2}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)=\mathrm{4} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\mathrm{4}+\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{4}\frac{\mathrm{1}}{\mathrm{6}}{ans} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

bah darun excellent...

$${bah}\:{darun}\:{excellent}... \\ $$

Commented by math1967 last updated on 10/Jan/19

dhanyabad

$${dhanyabad} \\ $$

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