Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.

Question

Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.

What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not help me anywhere. Perhaps I have to multiply something with $a$ only so that the expression becomes usable, what what to multiply?

Next, the expression we need to find is $\frac{3a^2 + 3a + 1}{a^3}$ , which I did not find any cool factorizations, and do not think anything will help here.

Can anyone help me?

Answer 1

Hint: As $$a=\frac{(\sqrt[3]{4}+\sqrt[3]{2}+1)(\sqrt[3]{2}-1)}{\sqrt[3]{2}-1}=\frac{1}{\sqrt[3]{2}-1} $$ $${\left(\frac{1}{a}+1\right)}^3=2$$ can you proceed ....

Answer 2

Note that \begin{eqnarray*} \frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{(a+1)^3-a^3}{a^3}. \end{eqnarray*} Note also that $a+1=\sqrt[3]{2}a$ so ... \begin{eqnarray*} \frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{2a^3-a^3}{a^3}=1. \end{eqnarray*}