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Answer:
[tex]\dfrac{2x\sqrt[4]{y^2}}{3}[/tex]
Step-by-step explanation:
We can first simplify under the radical, then take the 4th root.
[tex]\sqrt[4]{\dfrac{16x^{11}y^8}{3x^7y^6}}=\sqrt[4]{\dfrac{2^4x^{(11-7)}y^{(8-6)}}{3^4}}=\sqrt[4]{\dfrac{(2x)^4}{3^4}y^2}=\boxed{\dfrac{2x\sqrt[4]{y^2}}{3}}[/tex]
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The applicable rules of exponents are ...
(a^b)/(a^c) = a^(b-c)
(a^b)^c = a^(bc)
nth root is the same as an exponent of 1/n.
