Question Number 228331 by fantastic2 last updated on 07/Apr/26
$${x}=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:\&{x}−{y}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =? \\ $$
Answered by TonyCWX last updated on 07/Apr/26
$${x}\:=\:\frac{\sqrt{\mathrm{2}}\:+\:\mathrm{1}}{\:\sqrt{\mathrm{2}}\:−\:\mathrm{1}}\:=\:\frac{\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}\:−\:\mathrm{1}}\:=\:\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${y}\:=\:\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:−\:\mathrm{4}\sqrt{\mathrm{2}}\:=\:\mathrm{3}\:−\:\mathrm{2}\sqrt{\mathrm{2}}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$$ \\ $$$${x}\:−\:{y}\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{x}\:−\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\:=\:\mathrm{34} \\ $$$$\Rightarrow\:{x}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\mathrm{34}^{\mathrm{2}} \:−\:\mathrm{2}\:=\:\mathrm{1154} \\ $$$$\Rightarrow\:{x}^{\mathrm{4}} \:+\:{y}^{\mathrm{4}} \:=\:\mathrm{1154} \\ $$
Commented by fantastic2 last updated on 07/Apr/26
$${thanks} \\ $$