Can You Solve This Week's Math Challenge Equation?

[anemone]

Here is this week's POTW:
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Solve the equation \(\displaystyle x^3=4+\left\lfloor{x}\right\rfloor\).

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to IanCg for his correct solution, which you can find below::)

Spoiler

For $x < 0,\, x^3-4 < x-1 \leqslant\left\lfloor{x}\right\rfloor$ so there are no solutions in this range
For$ x >2,\, x^3 - 4 > x \geqslant\left\lfloor{x}\right\rfloor$ so no solutions in this range
for $0\le x < 1,\, x^3-4$ is negative and $\left\lfloor{x}\right\rfloor = 0$ so no solutions

View attachment 6303

The only range for a solution is $1\le x<2$ in this case $\left\lfloor{x}\right\rfloor = 1$
So the equation becomes $x^3 - 4 = 1 $
So the solution is $x = \left\lfloor{\sqrt[3]{5}}\right\rfloor$