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Question Number 132682 by otchereabdullai@gmail.com last updated on 15/Feb/21

$$\mathrm{expand}1-{\mathrm{x}}^8$$

Answered by MJS_new last updated on 15/Feb/21

$$1-x^8=(1-x^4)(1+x^4)=(1-x^2)(1+x^2)(1+x^4)=$$$$=(1-x)(1+x)(1+x^2)(1+x^4)$$

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

$$1-{\mathrm{x}}^8=1-{\left({\mathrm{x}}^4\right)}^2=(1-{\mathrm{x}}^4)(1+{\mathrm{x}}^4)=(1-{\mathrm{x}}^2)(1+{\mathrm{x}}^2)(1+{\mathrm{x}}^4)$$$$=(1-\mathrm{x})(1+\mathrm{x})(1+{\mathrm{x}}^2)({(1+{\mathrm{x}}^2)}^2-{2\mathrm{x}}^2)$$$$=(1-\mathrm{x})(1+\mathrm{x})(1+{\mathrm{x}}^2)({\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1)({\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1)$$

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