STD 11 - 5. linear inequalities — JEE STD 11 Science — Question

We have,

$\sqrt[3]{n+1}-\sqrt[3]{n} < \frac{1}{12}$

$\sqrt[3]{n+1} < \sqrt[3]{n}+\frac{1}{12}$

Cubing both sides, we get

$n+1 < n+3(n)^{23} \times \frac{1}{12}+3 \sqrt[3]{n} \times \frac{1}{144}+\frac{1}{1728}$

$\Rightarrow \quad 1 < \frac{3 n^{1 / 3}}{}-\left(n^{1 / 3}+\frac{1}{12}\right)+\frac{1}{1728}$

$\Rightarrow \quad n^{1 / 3}\left(n^{1 / 3}+\frac{1}{12}\right) > 1-\frac{1}{1728}$

$\Rightarrow \quad n^{1 / 3}\left(n^{1 / 3}+\frac{1}{12}\right) > \frac{1727}{432}$

Put $n=8$ only possible least positive integers.