expand-4-please-

Question Number 33515 by alekan251 last updated on 18/Apr/18

$${expand}\:\alpha^{\mathrm{4}} +\beta^{\beta\:\:} {please} \\ $$

Commented by math khazana by abdo last updated on 18/Apr/18

if you mean α^4 + β^4 from R = (α^2 )^2 +(β^2 )^2 = (α^2 +β^2 )^2 −2α^2 β^2 = (α^2 +β^2 −(√2) αβ)( α^2 +β^2 +(√2) αβ) if α andβ from R let fix β and the roots of α^2 −(√2) αβ +β^2 Δ = (−(√2)β)^2 −4β^2 = −2β^2 =(i(√2)β)^2 α_1 =(((√2)β +i(√2)β)/2) =(((√2)β)/2)(1+i) α_2 = (((√2) β −i(√2)β)/2) =(((√2)β)/2)(1−i) also we have α^2 +(√2)αβ +β^2 = α^2 −(√2)α(−β) +(−β)^2 so the roots are t_1 =−((β(√2))/2)(1+i) and t_2 =−((β(√2))/2)(1−i) α^4 +β^4 = (α −α_1 )(α−α_2 )(α −t_1 )(α −t_2 ) .

$${if}\:{you}\:{mean}\:\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\:\:\:\:{from}\:{R} \\ $$$$=\:\left(\alpha^{\mathrm{2}} \right)^{\mathrm{2}} \:+\left(\beta^{\mathrm{2}} \right)^{\mathrm{2}} \:=\:\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \\ $$$$=\:\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:\alpha\beta\right)\left(\:\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:\:\alpha\beta\right) \\ $$$${if}\:\alpha\:{and}\beta\:{from}\:{R}\:\:{let}\:{fix}\:\beta\:\:{and}\:{the}\:{roots}\:{of} \\ $$$$\alpha^{\mathrm{2}} \:\:−\sqrt{\mathrm{2}}\:\alpha\beta\:\:+\beta^{\mathrm{2}} \\ $$$$\Delta\:=\:\left(−\sqrt{\mathrm{2}}\beta\right)^{\mathrm{2}} \:−\mathrm{4}\beta^{\mathrm{2}} \:=\:−\mathrm{2}\beta^{\mathrm{2}} \:=\left({i}\sqrt{\mathrm{2}}\beta\right)^{\mathrm{2}} \\ $$$$\alpha_{\mathrm{1}} =\frac{\sqrt{\mathrm{2}}\beta\:+{i}\sqrt{\mathrm{2}}\beta}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}}\beta}{\mathrm{2}}\left(\mathrm{1}+{i}\right) \\ $$$$\alpha_{\mathrm{2}} =\:\frac{\sqrt{\mathrm{2}}\:\beta\:−{i}\sqrt{\mathrm{2}}\beta}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}}\beta}{\mathrm{2}}\left(\mathrm{1}−{i}\right)\:{also}\:{we}\:{have} \\ $$$$\alpha^{\mathrm{2}} \:\:+\sqrt{\mathrm{2}}\alpha\beta\:+\beta^{\mathrm{2}} \:=\:\alpha^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\alpha\left(−\beta\right)\:+\left(−\beta\right)^{\mathrm{2}} \:{so}\:{the} \\ $$$${roots}\:{are}\:\:{t}_{\mathrm{1}} =−\frac{\beta\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right)\:\:{and}\:{t}_{\mathrm{2}} =−\frac{\beta\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−{i}\right) \\ $$$$\alpha^{\mathrm{4}} \:+\beta^{\mathrm{4}} \:=\:\left(\alpha\:−\alpha_{\mathrm{1}} \right)\left(\alpha−\alpha_{\mathrm{2}} \right)\left(\alpha\:−{t}_{\mathrm{1}} \right)\left(\alpha\:−{t}_{\mathrm{2}} \right)\:. \\ $$

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