How to Simplify a Fraction with Cubed Roots in the Numerator and Denominator?

[Elena1]

Help to obtain a simple form of the following expression:

\(\displaystyle \frac{\sqrt[3]{a^2} -\sqrt[3]{ab}+\sqrt[3]{b}}{a+b}\)

 

[MarkFL]

Hello and welcome to MHB, Elena! :D

I have moved your thread here as this is a more appropriate place for your question. The "Introductions" sub-forum is where people may introduce themselves, by telling us a bit about themselves if they wish.

I have changed the thread title and edited the post so that the problem is in the body of the post, instead of the title.

Have I interpreted the problem correctly when I wrapped the code you provided with MATH tags?

In either case, can you show us what you have tried so far so we know where you are stuck and can provide better help?

 

[Elena1]

i need help to solve this exercise

 

[anemone]

Help to obtain a simple form of the following expression:

\(\displaystyle \frac{\sqrt[3]{a^2} -\sqrt[3]{ab}+\sqrt[3]{b}}{a+b}\)

Hi Elena, welcome to MHB!:)

I suspect the expression should read

\(\displaystyle \frac{\sqrt[3]{a^2} -\sqrt[3]{ab}+\sqrt[3]{b^2}}{a+b}\)

Could you please check and post back if my intuition is correct?

 

[Elena1]

yes it`s correct

 

[MarkFL]

I would suggest looking at the expansion:

\(\displaystyle \left(\sqrt[3]{a}+\sqrt[3]{b}\right)^3\)

to see if you can rewrite the denominator. :D

 

[anemone]

Perhaps you're not familiar with expanding a cube of a binomial?

I know this the cube root of $a$ and $b$ make the problem looks a bit intimidating, but you can try to make use of the substitution skill, such as what I would do below:

If we have to expand the cube of the binomial $x+y$, we see that

$\begin{align*}(x+y)^3&=(x+y)(x+y)(x+y)\\&=(x+y)(x^2+2xy+y^2)\\&=x(x^2+2xy+y^2)+y(x^2+2xy+y^2)\\&=x^3+2x^2y+xy^2+x^2y+2xy^2+y^3\\&=x^3+3x^2y+3xy^2+y^3\\&=x^3+3xy(x+y)+y^3\end{align*}$

Rewriting it to make $x^3+y^3$ the subject, and simplify the expression we get

$\begin{align*}x^3+y^3&=(x+y)^3-3xy(x+y)\\&=(x+y)((x+y)^2-3xy)\\&=(x+y)(x^2+2xy+y^2-3xy)\\&=(x+y)(x^2-xy+y^2)\end{align*}$

Now, if we let $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$, the equation above becomes

$(\sqrt[3]{a})^3+(\sqrt[3]{b})^3=(\sqrt[3]{a}+\sqrt[3]{b})((\sqrt[3]{a})^2-\sqrt[3]{a}\sqrt[3]{b}+(\sqrt[3]{a})^2)$

This is just

$a+b=(\sqrt[3]{a}+\sqrt[3]{b})((\sqrt[3]{a})^2-\sqrt[3]{a}\sqrt[3]{b}+(\sqrt[3]{a})^2)$

Now, can you proceed?

 

[Elena1]

thank you.after I was asked if I wrote correctly I said no, but actually I was wrong ... my mistake