Let N be the set of positive integers. For all $n \in N$, let ${f_n} = {\left( {n + 1} \right)^{\dfrac{1}{3}}} - {n^{\dfrac{1}{3}}}$ and $A = \left\{ {n \in N:{f_{n + 1}} < \dfrac{1}{{3{{\left( {n + 1} \right)}^{\dfrac{2}{3}}}}} < {f_n}} \right\}$, then
A) $A = N$
B) A is a finite set
C) The complement of A in N is nonempty
D) A and its complement in N are both infinite.

Hint:
We can draw the graph of the function $g\left( x \right) = {x^{\dfrac{1}{3}}}$ and take 3 points on the graph corresponding to 3 consecutive natural numbers $n,n + 1$ and $n + 2$ on the positive x axis. Then we can find the slope of the line joining two consecutive points. Then we can find the slope of the tangent at the middle point using derivatives. Then we can compare the slopes and form an inequality to find which values of N are in A. Thus, we can compare the options and get the required solution.

Complete step by step solution:
We are given that ${f_n} = {\left( {n + 1} \right)^{\dfrac{1}{3}}} - {n^{\dfrac{1}{3}}}$ …. (1)
Then, we can write ${f_{n + 1}} = {\left( {n + 2} \right)^{\dfrac{1}{3}}} - {\left( {n + 1} \right)^{\dfrac{1}{3}}}$ …… (2)
Now we can consider a function $g\left( x \right) = {x^{\dfrac{1}{3}}}$ which is defined for positive real numbers.
Now we can plot the graph of this function.

 We can take points A, B and C such that they are points on the graph corresponding to 3 consecutive natural numbers $n,n + 1$ and $n + 2$ on the positive x axis.
Now we can find the slope of the line joining AB.
We know that slope of the line joining the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
 $ \Rightarrow slope\,AB = \dfrac{{h\left( {n + 1} \right) - h\left( n \right)}}{{n + 1 - n}}$
On simplification, we get,
 $ \Rightarrow slope\,AB = \dfrac{{{{\left( {n + 1} \right)}^{\dfrac{1}{3}}} - {{\left( n \right)}^{\dfrac{1}{3}}}}}{1}$
So we have,
 $ \Rightarrow slope\,AB = {\left( {n + 1} \right)^{\dfrac{1}{3}}} - {\left( n \right)^{\dfrac{1}{3}}}$
On substituting equation (1), we get,
 $ \Rightarrow slope\,AB = {f_n}$ …… (3)
Similarly, we can find the slope of BC.
 $ \Rightarrow slope\,BC = \dfrac{{h\left( {n + 2} \right) - h\left( {n + 1} \right)}}{{\left( {n + 2} \right) - \left( {n + 1} \right)}}$
On simplification, we get,
 $ \Rightarrow slope\,BC = \dfrac{{{{\left( {n + 2} \right)}^{\dfrac{1}{3}}} - {{\left( {n + 1} \right)}^{\dfrac{1}{3}}}}}{1}$
So we have,
 $ \Rightarrow slope\,BC = {\left( {n + 2} \right)^{\dfrac{1}{3}}} - {\left( {n + 1} \right)^{\dfrac{1}{3}}}$
On substituting equation (2), we get,
 $ \Rightarrow slope\,BC = {f_{n + 1}}$ …. (4)
Now we can take the slope of the tangent at B, for that we find the derivative of the function.
 $ \Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{3}}}} \right)$
On differentiating we get,
 $ \Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{1}{{3{{\left( x \right)}^{\dfrac{2}{3}}}}}$
We need to find the slope at point B. so we can substitute the value of $x = n + 1$
 $ \Rightarrow slope\,at\,B = \dfrac{1}{{3{{\left( {n + 1} \right)}^{\dfrac{2}{3}}}}}$ ….. (5)
Since the curve of the function $h\left( x \right) = {\left( x \right)^{\dfrac{1}{3}}}$ is concave downwards, its slope decreases as x increases.
So, we can write the slopes as inequality,
 $ \Rightarrow slope\,BC\, < slope\,o\,tangent\,at\,B < slope\,AB$
On substituting (3), (4) and (5), we get,
 $ \Rightarrow {f_n} < \dfrac{1}{{3{{\left( {n + 1} \right)}^{\dfrac{2}{3}}}}} < {f_{n + 1}}$
This condition satisfies for any positive integer n. So, we can say that all the elements in A are exactly the same elements in N.
 $ \Rightarrow A = N$

Therefore, the correct answer is option A, $A = N$.

Note:
We must make sure that while drawing the graph of the function, we must draw with the function’s characteristics. We must draw only a graph which is concave downwards. The given function f is mapped from natural numbers to real numbers. So, the value of the function need not be a natural number. A is not finite as N is infinite. The complement of A in N is empty as $A = N$