Answer:
[tex]\large\boxed{\sqrt{xy^3}\cdot\sqrt[3]{x^2y}=\sqrt[6]{x^7y^{11}}}[/tex]
Step-by-step explanation:
[tex]\text{Use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\sqrt{xy^3}\cdot\sqrt[3]{x^2y}=(xy^3)^\frac{1}{2}(x^2y)^\frac{1}{3}\\\\\text{use}\ (ab)^n=a^nb^n\ \text{and}\ (a^n)^m=a^{nm}\\\\=x^\frac{1}{2}y^{(3)\left(\frac{1}{2}\right)}x^{(2)\left(\frac{1}{3}\right)}y^\frac{1}{3}\\\\\text{use}\ a^na^m=a^{n+m}\\\\=x^{\frac{1}{2}+\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{3}}\\\\\text{the common denominator is 6}[/tex]
[tex]\dfrac{1}{2}=\dfrac{1\cdot3}{2\cdot3}=\dfrac{3}{6}\\\\\dfrac{2}{3}=\dfrac{2\cdot2}{3\cdot2}=\dfrac{4}{6}\\\\\dfrac{3}{2}=\dfrac{3\cdot3}{2\cdot3}=\dfrac{9}{6}\\\\\dfrac{1}{3}=\dfrac{1\cdot2}{3\cdot2}=\dfrac{2}{6}\\\\x^{\frac{1}{2}+\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}=x^{\frac{3}{6}+\frac{4}{6}}y^{\frac{9}{6}+\frac{2}{6}}=x^{\frac{7}{6}}y^{\frac{11}{6}}=\sqrt[6]{x^7y^{11}}[/tex]
