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Question Number 172009 by Mikenice last updated on 23/Jun/22
solve:  if x=2017,y=2018 and z=2019,   find x^2 +y^2 +z^2 −xy−yz−zx
$${solve}: \\ $$$${if}\:{x}=\mathrm{2017},{y}=\mathrm{2018}\:{and}\:{z}=\mathrm{2019},\: \\ $$$${find}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx} \\ $$
Answered by som(math1967) last updated on 23/Jun/22
 x^2 +y^2 +z^2 −xy−yz−zx  =(1/2){(x−y)^2 +(y−z)^2 +(z−x)^2 }  =(1/2)×{(−1)^2 +(−1)^2 +2^2 }  =3
$$\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} −\boldsymbol{{xy}}−\boldsymbol{{yz}}−\boldsymbol{{zx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\boldsymbol{{x}}−\boldsymbol{{y}}\right)^{\mathrm{2}} +\left(\boldsymbol{{y}}−\boldsymbol{{z}}\right)^{\mathrm{2}} +\left(\boldsymbol{{z}}−\boldsymbol{{x}}\right)^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\left\{\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right\} \\ $$$$=\mathrm{3} \\ $$$$\: \\ $$

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