Find the least possible integer for which $\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$ Question
Find the least possible no for which $$\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$$
How do I reach a stage where I can deduce definitely the smallest integer value of $n$. I keep getting stuck after I cube on both sides and I still get cube root terms which I can't seem to simplify . Any help is appreciated.Thanks :)
 A: Let $n = x^3$
\begin{array}{c}
   \sqrt[3]{n+1} - \sqrt[3] n < \frac 1{12} \\
   \sqrt[3]{x^3+1} - x < \frac 1{12} \\
   \sqrt[3]{x^3+1} < x + \frac 1{12} \\
   x^3+1 < x^3 + \dfrac 14x^2 + \dfrac{1}{48}x + \dfrac{1}{1728} \\
   \dfrac 14x^2 + \dfrac{1}{48}x - \dfrac{1727}{1728} > 0 \\
   x^2 + \dfrac{1}{12}x - \dfrac{1727}{432} > 0 \\
   \left( x + \dfrac{1}{24} \right)^2 -  \dfrac{6911}{1728} > 0\\ 
   x + \dfrac{1}{24} >  \dfrac{\sqrt{6911}}{24\sqrt 3} \\ 
   x >  \dfrac{\sqrt{6911}}{24\sqrt 3} - \dfrac{1}{24} \\
   x > \dfrac{1}{72} (\sqrt{20733} - 3) \\
   x^3 > \left(\dfrac{1}{72} (\sqrt{20733} - 3)\right)^3 \\
   n > \dfrac{865 \sqrt{20733} - 7776}{15552} \\
   n > 7.5
\end{array}
The smallest value is $8$.
A: Since the function is convex, you have that: $$f(x+dx) < f(x) + f'(x)dx$$
$$dy <f'(x)dx$$
In this case we have:
$$dy  =\sqrt[3]{n+1}- \sqrt[3]{n} < \frac{1}{3}n^{-2/3}$$
We want the first $n$ such that $dy<1/12$.
If we compare both values we get that if (but not only if!) $n>8$, your condition holds. It is enough to check that $n=7$ does not satisfy the condition to find $n=8$.
A: HINT:
$$\sqrt[3]{n+1}-\sqrt[3] n=\dfrac{n+1-n}{(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}}$$
Now $3n^{2/3}<(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}<3(n+1)^{2/3}$
A: Here's a simple way to get an upper bound on the least $n$ without doing much in the way of algebra or numerical calculations:  
Note that the average value of the first $26$ values of $\sqrt[3]{n+1}-\sqrt[3]n$ is a telescoping sum,
$${(\sqrt[3]{27}-\sqrt[3]{26})+(\sqrt[3]{26}-\sqrt[3]{25})+\cdots+(\sqrt[3]{3}-\sqrt[3]{2})+(\sqrt[3]{2}-\sqrt[3]{1})\over26}={3-1\over26}={1\over13}$$
Since ${1\over13}\lt{1\over12}$, at least one of the numbers in the average must be less than $1\over12$.  So the number being sought is no greater than $26$.  
This only tells you where to look, of course.  If there is a simple way to zero in on the exact answer that doesn't use some combination of algebra and numerics along the lines of what's been done in other answers, I'd like to see it.
