<s>[INST] Algebraic manipulation with indices The question is:
For a>0 and $\sqrt{a}+\frac{1}{\sqrt{a}}=3$, find the value of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$
So I first squared the given equation and got:
$$a+\frac{1}{a}+2=9$$
$$a+\frac{1}{a}=7$$
Then to get the form of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$:
$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$$
And I just got stuck right here because I didn't really know what to do with the $\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$. So I looked into the solutions and apparently it's 
$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\sqrt{a}+\frac{1}{\sqrt{a}}$$
I'm not sure how those two are equal... 
 [/INST] $\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}=3$ !
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