(This is duplicated due do the lack of answers in the previous asking session,because I edited the unreadable math too long that this got lost in "recent questions".I was unable to delete the previous one.)
Variable here is "a" and "b".The question is to simplify the $\sqrt [3] { (\frac{a}{\sqrt{b^3} })\times {\frac {\sqrt{a^6b^2}}{a} }+ \frac{a}{b^2}}$
So these are my steps
=$\left(\frac {a^3b} {b^\frac {3} {2} }+ \frac {a} {b^2}\right)^\frac {1} {3}$
=$\left({a^3b^{1-\frac {3} {2}} } + \frac {a} {b^2}\right)^\frac {1} {3}$
=$\left({a^3b^\frac {-1} {2} } + \frac {a} {b^2}\right)^\frac {1} {3}$
Now the real question is,how to simplify this?Assuming the both fraction is simplified to "x" and "y",I get $(x+y)^\frac {1} {3}$ but my family said that it is expanded to this:
$(x+y)^\frac {1} {3}=x^\frac {1} {3}+y^\frac {1} {3}$ It should not be this because $(x+y)^2=x^2+2xy+y^2$
The thing is I noticed where $(x+y)^n$ have a pattern called pascal triangle.But this can be only applied when n≥1 where n is a integer.
The problem with the binomial theorem is that the answer is infinite,which I do not want to get this.I need to get an answer where it is not infinite and does not involve any roots.How do I simplify this?
I have checked with Issac Newton theory,but the Issac Newton theory is infinite.
I have searched online and I see is infinite answer: Link
(And by the way,even if I put it as a single fraction,$\left( \frac {a^3b^\frac {1} {2}+a} {b^2} \right)^\frac {1} {3}$,it still encounters the same problem.)
If this is impossible to simplify without avoiding infinite series,is there another method to solve the question,$\sqrt [3] { (\frac{a}{\sqrt{b^3} })\times {\frac {\sqrt{a^6b^2}}{a} }+ \frac{a}{b^2}}$?