factorize-inside-C-x-x-2-y-2-z-2-

Question Number 46882 by maxmathsup by imad last updated on 02/Nov/18

$${factorize}\:{inside}\:{C}\left[{x}\right]\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \\ $$

Commented by maxmathsup by imad last updated on 02/Nov/18

$${we}\:{have}\:\left({x}+{y}+{z}\right)^{\mathrm{2}\:} \:={x}^{\mathrm{2}\:} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \:+\mathrm{2}\left({xy}\:+{yz}\:+{zx}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \:=\left({x}+{y}+{z}\right)^{\mathrm{2}} \:−\left(\sqrt{\mathrm{2}\left({xy}\:+{yz}\:+{zx}\right.}\right)^{\mathrm{2}} \:{let}\:\:\mathrm{2}\left({xy}+\:{yz}\:+{zx}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \:=\left({x}+{y}+{z}\right)^{\mathrm{2}} \:−\left(\sqrt{{r}}{e}^{{i}\frac{\theta}{\mathrm{2}}} \right)^{\mathrm{2}} \: \\ $$$$=\left({x}+{y}+{z}\:−\sqrt{{r}}{e}^{{i}\frac{\theta}{\mathrm{2}}} \right)\left({x}+{y}+{z}\:+{r}\:{e}^{{i}\frac{\theta}{\mathrm{2}}} \right)\:. \\ $$

Commented by maxmathsup by imad last updated on 02/Nov/18

$${another}\:{method}\:{we}\:{have} \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \:={x}^{\mathrm{2}} \:−\left({i}\sqrt{{y}^{\mathrm{2}\:} +{z}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left({x}−{i}\sqrt{{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }\right)\left({x}+{i}\sqrt{{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }\right)\:{or} \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \:={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:−\left({iz}\right)^{\mathrm{2}} \:=\left(\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }−{iz}\right)\left(\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }+{iz}\right) \\ $$

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