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Applying some properties, this should be: $(\sqrt[n]{a} * \sqrt[k]{a} ) - (a^{\frac{n+k}{nk}})= 0$

But according to symbolab, not.

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I guess it's a symbolab error. But I would like to make sure.

ESCM
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    This is true if $a$ is real and positive, but not necessarily true if $a$ is complex. – eyeballfrog Mar 19 '18 at 14:33
  • @CarlosCampos An expression can't have solutions, it can have values. For positive $a$, the expression $\sqrt[p]{a}$ has a single value. However, the equation $x^p = a$ may have multiple solutions. – Arthur Mar 19 '18 at 14:52
  • @CarlosCampos Yes, it does. Because you have the solution right there. The solution is that $x$ has the value $\sqrt[p]{a}$, which is a single, well-defined value (as long as $a$ is a non-negative real number). – Arthur Mar 19 '18 at 14:58
  • @Arthur But same idea could be applied to complex numbers, isn't that the case? What do you mean with well-defined value? Thank you for your clarifications anyway. – Carlos Campos Mar 19 '18 at 15:06
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    @CarlosCampos If $a$ is a non-negative real (say an arbitrary complex number) and $p$ is a natural number, then $x^p = a$ has $p$ solutions, but none of them stand out as "better" than any other, so $\sqrt[p]{a}$ doesn't make sense, since there is no consistent way to pick a single one of the $p$ solutions that it can be. On the other hand, if $a$ is a positive real number, then $x^p = a$ still has $p$ solutions, but exactly one of those is positive, and therefore "nicer" than all the others. That is the one we pick to be $\sqrt[p]{a}$. – Arthur Mar 19 '18 at 15:14
  • @Arthur Really clear, thank you – Carlos Campos Mar 19 '18 at 15:17

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It is not a symbolab error.

If we have two powers $x^m$ and $x^n$, then $x^m\cdot x^n = x^{m+n}$. This is known as the Power Product Rule because we are taking the product of two powers, but with a common base $x$.

If $m = \dfrac{1}{i}$ and $n = \dfrac{1}{j}$ then the same is still applied, such that $x^{1/i}\cdot x^{1/j} = x^{1/i\,+\, 1/j}$.

When we raise $x$ to the power of a fraction with numerator $1$ and denominator $h$ (or the reciprocal of $h$), then this is the same as $\sqrt [h]x$. Therefore, this power rule also applies to roots (or radicals).

It follows, then, that symbolab makes no mistake in writing that $$\Large \left(\sqrt [n] a\cdot \sqrt[k] a\right) - \left(a^{(n+k)/nk}\right) = a^{1/n\, + \, 1/k} - a^{(n+k)/nk}.$$

But it is technically incorrect as it should equal $0$ since, $$\frac{n+k}{nk} = \frac{n}{nk} + \frac{k}{nk} = \frac{\require{\cancel}\cancel n}{\cancel nk} + \frac{\cancel k}{n\cancel k} = \frac 1k + \frac 1n.$$

Mr Pie
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