Question Number 50710 by Tawa1 last updated on 19/Dec/18
$$\mathrm{If}\:\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{15}\:\:\mathrm{and}\:\:\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}\:\:=\:\mathrm{85},\:\:\:\mathrm{find}\:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 19/Dec/18
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+{y}\right){z}+{z}^{\mathrm{2}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} +\mathrm{2}{xz}+\mathrm{2}{yz}+{z}^{\mathrm{2}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$=\mathrm{15}^{\mathrm{2}} −\mathrm{2}×\mathrm{85} \\ $$$$=\mathrm{55} \\ $$
Commented by Tawa1 last updated on 19/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by afachri last updated on 19/Dec/18
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} \:=\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} +\:\mathrm{2}\left(\boldsymbol{{xy}}+\boldsymbol{{yz}}+\boldsymbol{{xz}}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{15}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:=\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{85}\right) \\ $$$$\mathrm{225}−\mathrm{170}\:\:\:\:=\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} =\:\mathrm{55} \\ $$
Commented by Tawa1 last updated on 19/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$