solve in R x^3 −5=[x]


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Question Number 86426 by M±th+et£s last updated on 28/Mar/20

$s o l v e i n R$ $x^{3} - 5 = [x]$
	Answered by mr W last updated on 28/Mar/20

	$l e t x = n + δ$ ${(n + δ)}^{3} - 5 = [x] = n$ ${(n + δ)}^{3} - n = 5$ $w i t h n = 0, L H S < 1$ $w i t h n = 2, L H S ⩾ 6$ $\Rightarrow n = 1$ ${(1 + δ)}^{3} - 1 = 5$ ${(1 + δ)}^{3} = 6$ $δ = \sqrt[3]{6} - 1$ $\Rightarrow x = n + δ = \sqrt[3]{6} \approx 1.81712$
	Commented by M±th+et£s last updated on 28/Mar/20

	$g o d b l e s s y o u s i r$
	Answered by MJS last updated on 28/Mar/20

	$x^{3} = 5 + [x]$ $- 8 ⩽ x^{3} < - 1 \Rightarrow 5 + [x] = 3$ $- 1 ⩽ x^{3} < 0 \Rightarrow 5 + [x] = 4$ $0 ⩽ x^{3} < 1 \Rightarrow 5 + [x] = 5$ $1 ⩽ x^{3} < 8 \Rightarrow 5 + [x] = 6 \Rightarrow x^{3} = 6 \Leftrightarrow x = \sqrt[3]{6}$ $8 ⩽ x^{3} < 27 \Rightarrow 5 + [x] = 7$
	Commented by M±th+et£s last updated on 28/Mar/20

	$g o d b l e s s y o u s i r$
	Commented by M±th+et£s last updated on 28/Mar/20

	$s i r w h y - 8 ⩽ x < - 1$
	Commented by M±th+et£s last updated on 28/Mar/20

	$s o r r y x^{3}$
	Commented by MJS last updated on 28/Mar/20

	$- 8 ⩽ x^{3} < - 1 \Leftrightarrow - 2 ⩽ x < - 1 \Rightarrow [x] = - 2$ $generally$ $\forall a \in Z : a ⩽ x < a + 1 \Rightarrow [x] = a$ $and a ⩽ x < a + 1 \Leftrightarrow a^{3} ⩽ x^{3} < {(a + 1)}^{3}$
	Commented by M±th+et£s last updated on 28/Mar/20

	$t h a n k y o u s i r$
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