we are given
[tex] \sqrt[3]{\frac{8x^6y^9}{27y^3z^3}} [/tex]
we can split exponents
[tex] =\frac{\sqrt[3]{8}\sqrt[3]{x^6}\sqrt[3]{y^9}}{\sqrt[3]{27}\sqrt[3]{y^3}\sqrt[3]{z^3}} [/tex]
now, we can simplify it
and we get
[tex] =\frac{2x^2y^3}{3yz} [/tex]
now, we can see that y^3 is on top and y is on bottom
so, we can cancel it
[tex] =\frac{2x^2y^2}{3z} [/tex]
so, option-B..........Answer
