Solution of $(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$

Question

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.

Answer 1

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.

Answer 2

At least from what I see, we may have to complicate the question.. $$(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$$ $$\sqrt[3]{n+1}\cdot12-\sqrt[3]{n}\cdot12=1$$ $$\sqrt[3]{n+1}=k\Rightarrow n=k^3-1$$ $$-12\sqrt[3]{k^3-1}=1-k\cdot12$$ $$\left(-12\sqrt[3]{k^3-1}\right)^3=(1-k\cdot12)^3$$ $$-1728k^3+1728=1-36k+432k^2-1728k^3$$ $$36k-432k^2+1727=0$$ $$k=\dfrac{-36+\sqrt{36^2-4(-432)\cdot1727}}{2(-432)},k=\dfrac{-36-\sqrt{36^2-4(-432)\cdot1727}}{2(-432)}$$ $$k=-\dfrac{\sqrt{20733}-3}{72},k=\dfrac{3+\sqrt{20733}}{72}$$ Both solutions here are true after verification. $$\sqrt[3]{n+1}=-\dfrac{\sqrt{20733}-3}{72}, \sqrt[3]{n+1}=\dfrac{3+\sqrt{20733}}{72}$$ $$373248n+373248=186624-\sqrt{20733}\cdot20760,373248n+373248=186624+\sqrt{20733}\cdot20760$$ $$n=-\dfrac{865\cdot\sqrt{20733}+7776}{15552}, n=\dfrac{865\cdot\sqrt{20733}-7776}{15552}$$ Both solutions here are also true after verification.

Answer 3

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 $n=x^3$.

Answer 4

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:- $$f'(n)=\frac{f(n+h)-f(n)}{h}=\frac{(n+1)^{\frac{1}{3}}-n^{\frac{1}{3}}}{1}=\frac{1}{12}$$