<s>[INST] 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!
 [/INST] -----$\sqrt{9} = 3^2 = 3 * 3$
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:
$(a*a*a*a*a*a) \ne 2(a*a*a)$
$(a*a*a*a*a*a) = (a*a*a)*(a*a*a) = (a*a*a)^2$.
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