prove-that-x-8-x-6-x-3-x-1-gt-0-x-R-

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

p r o v e t h a t : x^{8} + x^{6} - x^{3} - x + 1 > 0, x \in R

Commented by Ari last updated on 15/Dec/21

thanks Mr!

Commented by mr W last updated on 15/Dec/21

= x^{8} - x^{4} + \frac{1}{4} + x^{6} - x^{3} + \frac{1}{4} + x^{4} - x^{2} + \frac{1}{4} + x^{2} - x + \frac{1}{4}

= {(x^{4} - \frac{1}{2})}^{2} + {(x^{3} - \frac{1}{2})}^{2} + {(x^{2} - \frac{1}{2})}^{2} + {(x - \frac{1}{2})}^{2}

> 0

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