anonymous
  • anonymous
3nth sqrt 50x^2z^5 X 3nth sqrt 15y^3
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
In this case, it's not so nice because you don't have things you can easily divide. But fractional exponents are valid notationally
anonymous
  • anonymous
So \[\sqrt[3]{a^2} = a^{\frac{2}{3}}\]
anonymous
  • anonymous
well idk if this is right but i got 5sqrt 6yz^2

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anonymous
  • anonymous
Wait, the original expression you have is \[\sqrt[3]{50x^2z^5} \times \sqrt[3]{15y^3}\] ?
anonymous
  • anonymous
yeah. because i broke 50 into radical 25 and and 2
anonymous
  • anonymous
Ok, so yes. If you bring the insides of both together you can factor out a 5 and a z.
anonymous
  • anonymous
So the only problem is that you should have a y and a z on the outside as well.
anonymous
  • anonymous
\[\sqrt[3]{50*15x^2y^3z^5} = \sqrt[3]{6*5^3x^2y^3z^3z^2}\] And we take out everything with a power of 3. \[= 5yz*\sqrt[3]{6x^2z^2}\]
anonymous
  • anonymous
ohh. okaiee. that makes sense. thanks once again
anonymous
  • anonymous
All of this is just remembering your rules for exponents when you multiply or raise to a power. a \[a^b*a^c = a^{b+c}\] \[(a^b)^c = a^{b*c}\]

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