Binomial Expansion without inifinty series 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
If this is impossible to simplify without avoiding infinite series,is there another method to solve the question,$\sqrt [3] { (\frac{a}{\sqrt{b^3} })\times {\frac {\sqrt{a^6b^2}}{a} }+ \frac{a}{b^2}}$?

In general,
$$(x+y)^{1/3}\ne x^{1/3}+y^{1/3}$$
For example, take $x=y=1$. If this weren't true, we'd have $\sqrt[3]2=2$.
In fact, if $x$ and $y$ are positive, we have:
$$(x+y)^{1/3}<x^{1/3}+y^{1/3}$$
