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Question Number 161295 by Ari last updated on 15/Dec/21

prove that:x^8 +x^6 −x^3 −x+1>0,x∈R

$${prove}\:{that}:{x}^{\mathrm{8}} +{x}^{\mathrm{6}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}>\mathrm{0},{x}\in{R} \\ $$

Commented byAri last updated on 15/Dec/21

thanks Mr!

Commented bymr W last updated on 15/Dec/21

=x^8 −x^4 +(1/4)+x^6 −x^3 +(1/4)+x^4 −x^2 +(1/4)+x^2 −x+(1/4) =(x^4 −(1/2))^2 +(x^3 −(1/2))^2 +(x^2 −(1/2))^2 +(x−(1/2))^2 >0

$$={x}^{\mathrm{8}} −{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}+{x}^{\mathrm{6}} −{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}+{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}+{x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $$=\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$ $$>\mathrm{0} \\ $$

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