Solution of $(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$ Solve the given equation for $n$
$(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$
How to approach this particular question? Sorry cannot show any work because the only approach I can see is take cube on both sides but that is complicating the equation.
 A: Let $x=\sqrt[3]{n+1}$ and $y=\sqrt[3]{n}$. Note that $(x-y)^3=x^3-y^3-3xy(x-y)=1-3xy(x-y)$. So
$$\frac{1}{12^3}=1-\frac{xy}{4}.$$
Solve for $xy$. We get $xy=a$, where $a$ is a mildly messy number.
We also have $x-y=b$, where $b=\frac{1}{12}$. 
So $(x+y)^2=b^2+4a$, and now we know that $x+y=\pm\sqrt{b^2+4a}$. 
We know $x+y$ and $x-y$, so we know $y$. Finally, $n=y^3$.
Remark: The strategy used here is the one that Cardano used to find the roots of a reduced cubic.
A: As ddsLeonardo commented, rewrite $$(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$$ as $$(n+1)^{1/3}=n^{1/3}+\frac{1}{12}$$ Cube both sides $$n+1=\left(n^{1/3}+\frac{1}{12}\right)^3=n+\frac{n^{2/3}}{4}+\frac{n^{1/3}}{48}+\frac{1}{1728}$$ Simplify to get $$n^{2/3}+\frac{n^{1/3}}{12}-\frac{1727}{432}=0$$ Define $x=n^{1/3}$ to get the quadratic $$x^2+\frac{x}{12}-\frac{1727}{432}=0$$ Solve for $x$ and then $n=x^3$.
A: Hint:
The equation we have involves differentiation of one function of $n$. That function is $f(n)=n^{\frac{1}{3}}$

According to the definition of differentiation:-
$$f'(n)=\frac{f(n+h)-f(n)}{h}=\frac{(n+1)^{\frac{1}{3}}-n^{\frac{1}{3}}}{1}=\frac{1}{12}$$
