Find approximate value of (a) in terms of (b) by expansion (b = cbrt[a+1])

[YehiaMedhat]

Given that $b=\sqrt[3]{a+1}$, such that b is an integer and a>0. An approximate value for $\sqrt[3]{a}$ can be obtained from the following series:
(a)$-1-1\frac{1}{3}b^3+...$

(b)$b-\frac{1}{3b^2}+...$

(c)$1-\frac{1}{3}b^3+...$

(d)$1-\frac{1}{3b^3}+...$
If you ask me about what I had in mind to solve this, I would just say none :') I don't have any idea about how this could be even touched!!. Simply, aren't (b) and (a) constants?!! so, how could I differentiate for getting the coefficients? It seems like a piece of fiction!

 

[Dr.Peterson]

Given that $b=\sqrt[3]{a+1}$, such that b is an integer and a>0. An approximate value for $\sqrt[3]{a}$ can be obtained from the following series:
(a)$-1-1\frac{1}{3}b^3+...$

(b)$b-\frac{1}{3b^2}+...$

(c)$1-\frac{1}{3}b^3+...$

(d)$1-\frac{1}{3b^3}+...$
If you ask me about what I had in mind to solve this, I would just say none :') I don't have any idea about how this could be even touched!!. Simply, aren't (b) and (a) constants?!! so, how could I differentiate for getting the coefficients? It seems like a piece of fiction!

At the least, you can tell us the context: What topics are you currently learning? Anything about series?

Then, since $b=\sqrt[3]{x}$ where $x=a+1$, maybe you can use whatever you've learned about series to approximate $\sqrt[3]{x}$ near $x=a$. Or, maybe you can solve for $\sqrt[3]{a}$ in terms of b, and approximate that function (for x near b). Or something ...

Give it a try. (I've just tossed those ideas out as possible first thoughts you might have to get started.)

 

[YehiaMedhat]

You mean I can expand by equaling $x=a+1\ \text {then}\ \sqrt[3]{x}=\sqrt[3]{a+1}\ \text {then I expand the cubic root for x?}$

 

[Dr.Peterson]

You mean I can expand by equaling $x=a+1\ \text {then}\ \sqrt[3]{x}=\sqrt[3]{a+1}\ \text {then I expand the cubic root for x?}$

Not really.

I was saying that there are several potential things you might try, so you should try working it out using any relevant function you can see. Trying and adjusting and trying again is an inherent part of a problem like this. So I expected you to make those attempts and report back on your successes and failures, rather than expect us to tell you exactly what to do. That's what it takes to learn problem solving.

My first suggestion was random, and probably will not lead to success. My second was closer, since I had given it a little more thought. Did you try solving for $\sqrt[3]{a}$ as a function of b?

I'll tell you another observation you might make: Most of the options involve $b^3$, so you might let $x=b^3$, so that you write $\sqrt[3]{a}=f(x)=f(b^3)$, and expand that as a series.