Rationalize the denominator. Assume that all expressions under radicals represent positive numbers.

Question

bekkah323 · Answer

$$\sqrt[3]{\frac{ 4 }{ 5x^5y^2 }}$$

anonymous · Answer

multiply top and bottom by $\sqrt[3]{25xy}$

bekkah323 · Answer

where did you get that?

anonymous · Answer

because $5\times 25=5^3$ and $\sqrt[3]{5^3}=5$

anonymous · Answer

similarly $x^5\times x=x^6$ and $\sqrt[3]{x^6}=x^2$

anonymous · Answer

and also $y^2\times y=y^3$ and $\sqrt[3]{y^3}=y$
so there will be no radical left in the denominator when you do this, just 
$$5x^2y$$

bekkah323 · Answer

Im sorry, i'm still a little confused on how you got that.

anonymous · Answer

ok lets go slow
your have to get rid of the cubed root in the denominator right?  i mean that is what "rationalize" means in this case

bekkah323 · Answer

ok, that part makes sense!

anonymous · Answer

so you do this by multiplying the denominator by whatever will make the inside stuff in to perfect cubes

bekkah323 · Answer

what do you mean perfect cubes?

anonymous · Answer

$y^3$ is a perfect cube
it is the cube of $y$

anonymous · Answer

$125$ is a perfect cube, it is the cube of $5$

bekkah323 · Answer

ahhh ok

anonymous · Answer

and finally $x^6$ is also a perfect cube, it is the cube of $x^2$

anonymous · Answer

now if you look above at the instructions i wrote, i hope the reasoning is more clear
if not, let me know