square root in algebra Have a question about roots. 
$$ \sqrt{9} = 3^2 = 3 \times 3 $$
$$ \sqrt{9a^6} = 3a^3  $$
But why does $ \sqrt{a^6} = a^3 $?
Is it correct vision?
$$ \sqrt{a^6} = \sqrt{a\times a \times a \times a \times a \times a} = \sqrt{2(a \times a \times a)}   $$
Can anybody add more details about my question? 
Thanks!
 A: -----$\sqrt{9} = 3^2 = 3 * 3$
Incorrect.
$\sqrt{9} = \sqrt{3^2} = 3$
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----but why $\sqrt{a^6}=a^3$
Because $\sqrt{a^6} = \sqrt{(a^3)^2} = a^3$. 
[Assuming $a \ge 0$.  If $a < 0$ then $a^6 = |a|^6 =|a^6|  > 0$ and $\sqrt{a^6} = |a|^3$.  For the rest of the post I'm assuming $a \ge 0$.]
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----Is it correct vision ?
$\sqrt{a^6} = \sqrt{a∗a∗a∗a∗a∗a}=\sqrt{2(a∗a∗a)}$
Not quite.  $\sqrt{a^6} = \sqrt{a∗a∗a∗a∗a∗a}=\sqrt{(a∗a∗a)*(a*a*a)}=\sqrt{a^3*a^3} = \sqrt{(a^3)^2} = a^3$
Note:
$(a*a*a*a*a*a) \ne 2(a*a*a)$
$(a*a*a*a*a*a) = (a*a*a)*(a*a*a) = (a*a*a)^2$.
A: In your vision, you grouped $a*a*a$ together and multiplied. But you forgot that, $x*x=x^2\text{ and}\neq 2x$.
So, even in your groupings
$(a*a*a)*(a*a*a)=(a*a*a)^2$
To your originial question, If you are ever in doubt remember, $\sqrt x$ is same as saying $x^\frac{1}{2}$. 
So, if you put $x=9a^6$ you will get, $(9a^6)^\frac{1}{2}$. 
The exponent will distribute itself over the bases, using this identity $(xy)^n=x^n\times y^n$. 
So overall, the expression will reduce to
$(9)^\frac{1}{2}\times(a^6)^\frac{1}{2}$
Now, comes a third identity $(x^m)^n=x^{m\times n}$. So, $(9)^\frac{1}{2} = (3^2)^\frac{1}{2}=3^{{2}\times\frac{1}{2}}=3$.
A: It is not a correct vision, the correct one would be:
$$\sqrt{a^6}=\sqrt{a\times a\times a\times a\times a\times a}=\sqrt{(a\times a\times a)^2}=\vert a\times a\times a \vert=\vert a^3\vert.$$
