Solve-x-3-x-3-x-R-

Question Number 108741 by mr W last updated on 18/Aug/20

S o l v e x^{3} - [x] = 3

(x \in R)

Answered by Sarah85 last updated on 18/Aug/20

first, I think 1 < x < 2

\Rightarrow [x] = 1 \Rightarrow x = \sqrt[3]{4}

second, proof :

x = z + q; z \in Z \land q \in Q \land 0 ⩽ q < 1

{(z + q)}^{3} - z = 3

q = - z + \sqrt[3]{z + 3}

0 ⩽ q < 1 \Leftrightarrow z ⩽ \sqrt[3]{z + 3} < z + 1

z \in Z \Rightarrow z = 1 \Rightarrow 1 < x < 2 \dots

Commented by ajfour last updated on 18/Aug/20

e x c e l l e n t w a y, t h a n k s .

Commented by Sarah85 last updated on 18/Aug/20

thank you

Commented by Sarah85 last updated on 18/Aug/20

for n \in Z ∖ {0} the equation x^{3} - [x] = n has only

one solution \Leftrightarrow {there}^{'} s always only one possible

value for z satisfying z ⩽ \sqrt[3]{z + n} < z + 1

Commented by mr W last updated on 19/Aug/20

t h a n k s!

Answered by mr W last updated on 19/Aug/20

s a y [x] = n

\Rightarrow x = n + f w i t h 0 ⩽ f < 1

{(n + f)}^{3} - n = 3

n + 3 = {(n + f)}^{3} ⩾ n^{3}

(n - 1) n (n + 1) ⩽ 3

\Rightarrow n = 0^{*} o r

\Rightarrow n = 1

{(1 + f)}^{3} - 1 = 3

\Rightarrow {(1 + f)}^{3} = 4

\Rightarrow f = \sqrt[3]{4} - 1

\Rightarrow x = 1 + \sqrt[3]{4} - 1 = \sqrt[3]{4}

^{*} w i t h n = 0 :

f^{3} = 3 \Rightarrow f = \sqrt[3]{3} > 1 \Rightarrow n o s o l u t i o n

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