Solve the equation `x^(3)-[x]=3`, where `[x]` denotes the greatest integer less than or equal to `x`.

Explanation

We have `x^(3)-[x]=3`
`impliesx^(3)-3=[x]`
Let `f(x)=x^(3)-3` and `g(x)=[x]`
It is clear from the graphs, the point of intersection of two curves `y=f(x)` and `y-g(x)` lies between (1,0) and (2,0).

`:.1ltxlt2`
We have `f(x)=x^(3)-3` and `g(x)=1`
or `x^(3)-3=1impliesx^(3)=4`
`:.x=(4)^(1//3)`
Hence `x=4^(1//3)` is the solution of the equation `x^(3)-[x]=3`
Aliter
`:'x=[x]+f,0leflt1`
Then given equation reduces to
`x^(3)-(x-f)=3impliesx^(3)-x=3-f`
Hence it follows that
`2ltx^(3)-xle3`
`implies2ltx(x+1)(x-1)le3`
Further for `xge2`, we have `x(x+1)(x-1)ge6lt3`
For `xlt-1`, we have `x(x+1)(x-1)lt0lt2`
For `x=-1`, we have `x(x+1)(x-1)=0lt2`
For `-1ltxle0`, we have `x(x+1)(x-1)le-xlt1`
adn for `0ltxle1`, we have `x(x+1)(x-1)ltxltx^(3)le1`
Therefore `x` must be `1ltxlt2`
`:.[x]=1`
Now the original equation can be written as
`x^(3)-1=3impliesx^(3)=4`
Hence `x=4^(1//3)` is the solution of the given equation.