Question

Please help answer quickly!

Answer 1

The fourth root of 16/81 is 2/3.

x^11/x^7 = x^4. The fourth root of x^4 is x.

y^8/y^6 = y^2 which remains inside the fourth root.

Answer: B.

Answer 2

$\sqrt[4]{\dfrac{16x^{11}y^8}{81x^7y^6}}=\sqrt[4]{\dfrac{16}{81}}\cdot\sqrt[4]{\dfrac{x^{11}}{x^7}}\cdot\sqrt[4]{\dfrac{y^8}{y^6}}=\dfrac{\sqrt[4]{16}}{\sqrt[4]{81}}\cdot\sqrt[4]{x^{11-7}}\cdot\sqrt[4]{y^{8-6}}\\\\=\dfrac{2}{3}\cdot\sqrt[4]{x^4}\cdot\sqrt[4]{y^2}=\dfrac{2}{3}\cdot x\cdot y^\frac{2}{4}=\dfrac{2}{3}x\cdot y^\frac{1}{2}=\dfrac{2}{3}x\sqrt{y}$

$Used:\\\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\\sqrt[n]{a^n}=a\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}$

But your answer is:

$...=\dfrac{2}{3}\cdot\sqrt[4]{x^4}\cdot\sqrt[4]{y^2}=\dfrac{2x\left(\sqrt[4]{y^2}\right)}{3}$