Question Number 132682 by otchereabdullai@gmail.com last updated on 15/Feb/21
$$\mathrm{expand}\:\mathrm{1}−\mathrm{x}^{\mathrm{8}} \\ $$
Answered by MJS_new last updated on 15/Feb/21
$$\mathrm{1}−{x}^{\mathrm{8}} =\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)= \\ $$$$=\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right) \\ $$
Commented by otchereabdullai@gmail.com last updated on 15/Feb/21
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{u}\:\mathrm{prof}\:\mathrm{mjs} \\ $$
Answered by mathmax by abdo last updated on 16/Feb/21
$$\mathrm{1}−\mathrm{x}^{\mathrm{8}} \:=\mathrm{1}−\left(\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} \:=\left(\mathrm{1}−\mathrm{x}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)=\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right) \\ $$$$=\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} \right) \\ $$$$=\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\mathrm{x}\:+\mathrm{1}\right) \\ $$