1
$\begingroup$

I'm in a intermediate algebra class and am confused about how to get the simplified form of $\sqrt[3]{x^{10}}$

I tend to want to write it as $x^{10/3}$ creating a mixed fraction then simplifying that to get $x^{3}\sqrt[3]{x}$

However, when asking a friend they explained that if we look at it by going $\sqrt[3]{x^{8}}\sqrt[3]{x^{2}}$ then would get $x^2\sqrt[3]{x^{2}}$

Could someone please help with which one is correct and if the top one is correct explain why.

$\endgroup$
5
  • $\begingroup$ Please fix the title - it is a bit misleading. $\endgroup$
    – AD - Stop Putin -
    May 24, 2012 at 19:43
  • 1
    $\begingroup$ We have $x^a\cdot x^b= x^{a+b}$ and $(x^a)^b=x^{ab}$. $\endgroup$
    – AD - Stop Putin -
    May 24, 2012 at 19:45
  • $\begingroup$ @AD.feel free to change the title. I also understand exponent rules but for some reason the radical is throwing me off $\endgroup$
    – Brooke.
    May 24, 2012 at 19:47
  • $\begingroup$ BTW welcome to math.SE $\endgroup$
    – AD - Stop Putin -
    May 24, 2012 at 19:56
  • $\begingroup$ "Simplify" is a term that cannot be defined precisely. A "simplification" that is best for one purpose is not necessarily best for another. I think that $x^{10/3}$ is a good general purpose simplification. But in a class, what is best is effectively what teacher thinks best. $\endgroup$
    – André Nicolas
    May 24, 2012 at 21:12

3 Answers 3

Reset to default
5
$\begingroup$

$x^2\sqrt[3]{x^2}=x^2\cdot x^{2/3}=x^{8/3}\neq x^{10/3}$ ...

The mistake is at $\,\,\sqrt[3]{x^8}\sqrt[3]{x^2}\neq x^2\sqrt[3]{x^2}\,\,$ since $\,\,\sqrt[3]{x^8}\neq x^2\,\,$ but $\,x^{8/3}$

$\endgroup$
2
  • 2
    $\begingroup$ so $x^{3}\sqrt[3]{x}$ is correct? $\endgroup$
    – Brooke.
    May 24, 2012 at 19:51
  • $\begingroup$ Yes it is! $ \,\,\,\, $ $\endgroup$
    – AD - Stop Putin -
    May 24, 2012 at 19:54
3
$\begingroup$

We have $$x^{10/3}=x^{(9 + 1)/3}=x^{9/3 + 1/3}=x^{3 + 1/3}=x^3x^{1/3}$$

$\endgroup$
0
$\begingroup$

you can try $x^6*x^4$,we in your case it would be $x^2*x^{4/3}$,there are many forms,for example as @AD. indicated

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .