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if-x-2-y-2-z-2-14-2-x-2y-3z-find-T-xyz-x-3-y-3-z-3-




Question Number 190131 by HeferH last updated on 28/Mar/23
 if:  x^2 +y^2 +z^2  + 14 = 2(x + 2y + 3z)   find:  T=((xyz)/(x^3 +y^3 +z^3 ))
$$\:\mathrm{if}:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{14}\:=\:\mathrm{2}\left(\mathrm{x}\:+\:\mathrm{2y}\:+\:\mathrm{3z}\right) \\ $$$$\:\mathrm{find}:\:\:\mathrm{T}=\frac{\mathrm{xyz}}{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{z}^{\mathrm{3}} }\: \\ $$
Answered by som(math1967) last updated on 28/Mar/23
 x^2 +y^2 +z^2 +14−2x−4y−6z=0   (x^2 −2x+1)+(y^2 −4y+4)+(z^2 −6z+9)=0  (x−1)^2 +(y−2)^2 +(z−3)^2 =0  if x,y,z are real then  (x−1)^2 =0⇒x=1  (y−2)^2 =0⇒y=2  (z−3)^2 =0 ⇒z=3   T=((xyz)/(x^3 +y^3 +z^3 ))=((1.2.3)/(1+8+27))=(6/(36))=(1/6)
$$\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{14}−\mathrm{2}{x}−\mathrm{4}{y}−\mathrm{6}{z}=\mathrm{0} \\ $$$$\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)+\left({y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{4}\right)+\left({z}^{\mathrm{2}} −\mathrm{6}{z}+\mathrm{9}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} +\left({z}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${if}\:{x},{y},{z}\:{are}\:{real}\:{then} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$$\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{y}=\mathrm{2} \\ $$$$\left({z}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{z}=\mathrm{3} \\ $$$$\:{T}=\frac{{xyz}}{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} }=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}}{\mathrm{1}+\mathrm{8}+\mathrm{27}}=\frac{\mathrm{6}}{\mathrm{36}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by HeferH last updated on 28/Mar/23
thank you :)
$$\left.\mathrm{thank}\:\mathrm{you}\::\right) \\ $$

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