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

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