MHB Why Are Square Roots of Cubes Not Always Equal?

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The discussion centers on the confusion surrounding the equality of square roots and cube roots, specifically why \(\sqrt[3]{(1+x^{3})^{2}}=(1+x^{3})^{\frac{2}{3}}\) holds true while \(\sqrt{(x-2)^{3}}\neq (x-2)^{\frac{3}{2}}\). It is clarified that the latter equality is only valid when \((x-2)^3 \geq 0\), meaning it is not defined for all real numbers. The conversation also touches on the implications of negative values in exponentiation, noting that operations involving square roots are typically defined for nonnegative numbers. The distinction between cube roots, which are defined for all real numbers, and square roots is emphasized as a key factor in understanding these equations. Overall, the discussion highlights the importance of considering the domain of the expressions involved.
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Hi,

I have a very basic question that suddenly hit me regarding square roots.

Why this is equal
\[\sqrt[3]{(1+x^{3})^{2}}=(1+x^{3})^{^{\frac{2}{3}}}\]

but this isn't

\[\sqrt{(x-2)^{3}}\neq (x-2)^{\frac{3}{2}}\]

(well according to Maple it isn't)

I understand why the first one is correct, but I assumed to believe that also the second one is equal and now I am confused.
 
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Yankel said:
Hi,

I have a very basic question that suddenly hit me regarding square roots.

Why this is equal
\[\sqrt[3]{(1+x^{3})^{2}}=(1+x^{3})^{^{\frac{2}{3}}}\]

but this isn't

\[\sqrt{(x-2)^{3}}\neq (x-2)^{\frac{3}{2}}\]

(well according to Maple it isn't)

I understand why the first one is correct, but I assumed to believe that also the second one is equal and now I am confused.
Wolfram Alpha has no problem with it. (Note though that there is an issue when x - 2 < 0. I don't know why.)

-Dan
 
They are equal when $(x-2)^3 \geq 0$, but that is not defined for all $x \in \mathbb{R}$.

Consider $\sqrt{(-1)^6}$. What is the result of this operation? If you work inside out, you'll get $$\sqrt{(-1)^6} = \sqrt{1} = 1.$$ On the other hand, if you apply the exponents rule, you get $$\sqrt{(-1)^6} = (-1)^{\frac{6}{2}} = (-1)^3 = -1.$$ Is mathematics contradicting itself? Could our whole world be CRUMBLING BEFORE THE MIGHT OF EXPONENTIATION? Not really. The subtlety is that the operations are defined for nonnegative real numbers, letting the theory work smoothly. When we take in account negative real numbers as well, we take the order exponentiation - root to enable such operations.

In some cases it is not even possible to do so: in the real numbers there is no thing as $\sqrt{(-1)^5}$ because it is not defined.

Hope this has helped. Cheers! :D
 
topsquark said:
(Note though that there is an issue when x - 2 < 0. I don't know why.)

-Dan

What would a calculator make of $(-1)^{0.66667}$?

Oh, and my favorite:
$$-1=(-1)^{\frac 23 \cdot \frac 32}=((-1)^{\frac 23})^{\frac 32}=1^{\frac 32}=1$$

EDIT: Ah, Fantini was quicker than me!
 
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Thank you, but if the issue here is the expression under the square root being positive or negative, then how come the first expression is equal ?

1+x^3 is not positive for every x in R, and yet, Maple seem to think it's Ok.
 
You used the right term: square root. What you have first is a cubic root, which is defined for all real numbers. :D
 
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