Преобразуйте данное выражение к виду n√А 1)√2*³√32)⁴√2*⁶√33)⁴√3b³ *√3b

Решите задачу:

$1) \ \sqrt{2}\sqrt[3]{3} = 2^\frac{1}{2}\cdot3^\frac{1}{3} = 2^\frac{1*3}{2*3}\cdot 3^\frac{1*2}{3*2} = 2^\frac{3}{6}\cdot3^\frac{2}{6} = (2^3)^\frac{1}{6}\cdot(3^2)^\frac{1}{6} =\\\\= (2^3\cdot3^2)^\frac{1}{6} = (8\cdot9)^\frac{1}{6} = \boxed{\sqrt[6]{72}}$

$2) \ \sqrt[4]{2}\sqrt[6]{3} = 2^\frac{1}{4}\cdot3^\frac{1}{6} = 2^\frac{1*3}{4*3}\cdot3^\frac{1*2}{6*2} = 2^\frac{3}{12}\cdot3^\frac{2}{12} = (2^3)^\frac{1}{12}\cdot(3^2)^\frac{1}{12} =\\\\= (2^3\cdot3^2)^\frac{1}{12} =(8\cdot9)^\frac{1}{12} = \boxed{\sqrt[12]{72}}$

$3) \ \sqrt[4]{3b^3}\sqrt{3b} = (3b^3)^\frac{1}{4}\cdot(3b)^\frac{1}{2} = (3b^3)^\frac{1}{4} \cdot(3b)^\frac{2}{4} = (3b^3)^\frac{1}{4}\cdot(9b^2)^\frac{1}{4} =\\\\= (27b^5)^\frac{1}{4} = \boxed{\sqrt[4]{27b^5}}$