The least positive integer n for which sqrt[3 | vedclass

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(C) Let $f(n) = \sqrt[3]{n+1} - \sqrt[3]{n}$.Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,we have:$\sqrt[3]{n+1} - \sqrt[3]{n} = \frac{(n+1)

Vedclass · Accepted Answer

$8$

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(C) Let $f(n) = \sqrt[3]{n+1} - \sqrt[3]{n}$.Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,we have:$\sqrt[3]{n+1} - \sqrt[3]{n} = \frac{(n+1) - n}{(n+1)^{2/3} + (n+1)^{1/3}n^{1/3} + n^{2/3}} = \frac{1}{(n+1)^{2/3} + (n+1)^{1/3}n^{1/3} + n^{2/3}}$.We want $\frac{1}{(n+1)^{2/3} + (n+1)^{1/3}n^{1/3} + n^{2/3}}  12$.For large $n$,$(n+1)^{2/3} \approx n^{2/3}$,so the expression is approximately $3n^{2/3}$.Setting $3n^{2/3} \approx 12$,we get $n^{2/3} \approx 4$,so $n \approx 4^{3/2} = 8$.Testing $n=7$: $\sqrt[3]{8} - \sqrt[3]{7} = 2 - 1.9129 = 0.0871$,and $\frac{1}{12} \approx 0.0833$. Since $0.0871 > 0.0833$,$n=7$ does not satisfy the inequality.Testing $n=8$: $\sqrt[3]{9} - \sqrt[3]{8} = 2.08008 - 2 = 0.08008$. Since $0.08008 Thus,the least positive integer is $8$.

The least positive integer $n$ for which $\sqrt[3]{n+1}-\sqrt[3]{n} < \frac{1}{12}$ is

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