<s>[INST] 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 :)
 [/INST] 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 $8$.
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