Since $\sqrt{x^6}=\sqrt{(x^3)^2}$, wouldn't $\sqrt{(x^3)^2}=|x^3|$?
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2You are right, $\sqrt{x^6}=|x^3|$. – Wojowu Nov 30 '16 at 20:59
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2Is there some more context? Where did you find $\sqrt{x^6}=x^3$? It is generally true that $\sqrt{x^6}=|x^3|$ as you write. – Milo Brandt Nov 30 '16 at 20:59
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http://imgur.com/w1sovub – Allison Nov 30 '16 at 21:08
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1It does say "for any nonnegative real number $a$" in that image. Presumably, the remainder should also be read in the context of nonnegative reals. – Magdiragdag Nov 30 '16 at 21:23
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1And do have a look at first bolded line in that image where it says "we assume no radicals involve negative quantities raised to even powers". – Magdiragdag Nov 30 '16 at 21:26
