Let α be a root of x^5 −x^3 +x−2=0 Then prove that [α^6 ]=3 where[λ] denotes greatest integer less than or equal λ


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Question Number 121099 by Anuragkar last updated on 05/Nov/20

$L e t α b e a r o o t o f x^{5} - x^{3} + x - 2 = 0$ $T h e n p r o v e t h a t [α^{6}] = 3 w h e r e [λ] d e n o t e s g r e a t e s t i n t e g e r$ $l e s s t h a n o r e q u a l λ$
	Answered by TANMAY PANACEA last updated on 05/Nov/20

	$f (x) = x^{5} - x^{3} + x - 2$ $f (0) < 0$ $f (1) < 0$ $f (2) > 0$ $s o r o o t 2 > α > 1$ $u s i n g g r a p h a p p . . α = 1.206 \to α^{6} \approx 3.08$ $[α^{6}]$ $= [3.08] = 3$
	Commented by TANMAY PANACEA last updated on 05/Nov/20

	Commented by Anuragkar last updated on 13/Nov/20

	$g o o d a p p r o a c h . . . . . b u t t h e m a t t e r o f f a c t i s t h a t i s h a s b e c o m e$ $t o o m u c h c a l c u l a t i o n b a s e d . . . . i n a n e x a m h a l l$ $u w i l l n o t c a l c u l a t e s u c h h u g e v a l u e s . . .$ $I s u g g e s t u t o t a k e h e l p o f s o m e i n e q u a l i t i e s$ $t o f i n d t h e r a n g e o f α$
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