MHB Find the value of 3/a + 3/a² + 3/a³.

[anemone]

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

 

[Prove It]

Re: Find the value of 3/a+3/a²+3/a³.

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

Am I missing something, or can we just substitute the value of a? Or are you expecting this to simplify to something simple? :P

 

[anemone]

Re: Find the value of 3/a+3/a²+3/a³.

Am I missing something, or can we just substitute the value of a? Or are you expecting this to simplify to something simple? :P

Hi Prove It, I really think we have to substitute the value of a into the intended expression and further simplifying it from there...:)

 

[Opalg]

Re: Find the value of 3/a+3/a²+3/a³.

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

Let $\lambda = \sqrt[3]2$, so that $\lambda^3 = 2$. Then $(\lambda-1)a = (\lambda-1)(\lambda^2 + \lambda +1) = \lambda^3-1 = 2-1=1.$ Therefore $1/a = \lambda-1$, and $$\begin{aligned}\frac{3}{a}+\frac{3}{a^2}+\frac{3}{a^3} &= 3\bigl((\lambda-1) + (\lambda-1)^2 + (\lambda-1)^3\bigr) \\ &= 3(\lambda-1)(\lambda^2 - \lambda + 1) \\ &= 3(\lambda^3 -2\lambda^2 +2\lambda - 1) \\ &= 3(1+2\lambda-2\lambda^2) \\ &= 3 + 6\sqrt[3]2 - 6\sqrt[3]4.\end{aligned}$$ I don't see that it can be simplified further than that.

 

[anemone]

Re: Find the value of 3/a+3/a²+3/a³.

Thanks Opalg for participating in this problem and my answer is 'quite' similar to yours too.:)

My solution:

I noticed that $$a=\sqrt[3]{4}+\sqrt[3]{2}+1=1+\sqrt[3]{2}+\sqrt[3]{4}$$ is actually the sum of the first three terms of a geometric progression with first term and common ratio to be 1 and $$\sqrt[3]{2}$$ respectively and I found another way to rewrite $$a$$, i.e.

$$a=S_3=\frac{1((\sqrt[3]{2})^3-1)}{\sqrt[3]{2}-1}=\frac{1}{\sqrt[3]{2}-1}$$.

Therefore, the intended expression could be found by substituting this formula for $$a$$ into it to get:

$$\frac{3}{a}+\frac{3}{a^2}+\frac{3}{a^3}=\frac{3}{a^3}\left(1+a+a^2\right)=3(\sqrt[3]{2}-1)((\sqrt[3]{2})^2-\sqrt[3]{2}+1)$$

My little note to Prove It:

I am truly sorry, Prove It for telling you to substitute the value of $$a$$ straight into the equation and then to do the simplification, because what I am posting now is totally going against what I said to you but on the level, I did try to approach it using two ways and the initial one was to substitute first and simplify next.