sqrt3(x^2/yz^5)z

Is this your intended question Alisson?

$$\\\sqrt3\times \dfrac{x^2}{yz^5}\times z\\\\\\
=\sqrt3\times \dfrac{x^2}{yz^4}\\\\\\
=\dfrac{\sqrt{3}\;x^2}{yz^4}\\\\\\$$

[ x² / (yz⁵) ]^(1/3) · z  

To rationalize the denominator under the radical sign, multiply the numerator and the denominator by y²z:

[ x²·y²z / (yz⁵·y²z) ]^(1/3) · z  

=  [ x²·y²z / (y³z⁶) ]^(1/3) · z 

=  (x²·y²z)^1/3 / (yz²)  · z 

=  (x²·y²z)^1/3 / (yz)

what happened with the squart3

or do you mean

$$\sqrt[3]{x^2/yz^5)}\times z\\\\\\
=\sqrt[3]{\dfrac{x^2}{yz^5}}\times z\\\\\\
=\dfrac{x^{2/3}}{y^{1/3}z^{5/3}}\times z\\\\\\
=\dfrac{x^{2/3}}{y^{1/3}z^{3/3}z^{2/3}}\times z\\\\\\
=\dfrac{x^{2/3}}{y^{1/3}z^{2/3}}\\\\\\
=\sqrt[3]{\dfrac{x^{2}}{yz^{2}}}\\\\\\$$

I think Geno's answer is another presentation of this.   I am sure it is correct to.