Certainly! Here you go:
1. First, we rewrite the expression in terms of its components:
[tex]\[
3 \cdot x^2 \cdot y^2 \cdot \sqrt[4]{7 \cdot x \cdot y^3}
\][/tex]
2. Next, we rewrite the radicals using properties of rational exponents:
[tex]\[
3 x^2 y^2 \sqrt[4]{7 x y^3}
\][/tex]
3. Then, we express each component under the root as a product of terms with rational exponents:
[tex]\[
3^1 \cdot 7^{\frac{1}{4}} \cdot x^2 \cdot x^{\frac{1}{4}} \cdot y^2 \cdot y^{\frac{3}{4}}
\][/tex]
4. We combine the like terms in the expression:
[tex]\[
(3^4)^{\frac{1}{4}} \cdot 7^{\frac{1}{4}} \cdot x^{2 + \frac{1}{4}} \cdot y^{2 + \frac{3}{4}}
\][/tex]
5. Finally, we simplify the exponents and terms:
[tex]\[
(81)^{\frac{1}{4}} \cdot(7)^{\frac{1}{4}} \cdot x^{\frac{9}{4}} \cdot y^{\frac{11}{4}}
\][/tex]
These are the steps for simplifying the expression using the properties of rational exponents:
1. [tex]$3 \cdot x^2 \cdot y^2 \cdot\left(7\right.$[/tex] 故 [tex]$\left.\cdot x^{\frac{1}{4}} \cdot y^{\frac{3}{4}}\right)$[/tex]
2. [tex]$3 x^2 y^2 \sqrt[4]{7 x y^3}$[/tex]
3. [tex]$3^1 \cdot 7^{\frac{1}{4}} \cdot x^2 \cdot x^{\frac{1}{4}} \cdot y^2 \cdot y^{\frac{3}{4}}$[/tex]
4. [tex]$\left(3^4\right)^{\frac{1}{4}} \cdot 7^{\frac{1}{4}} \cdot x^{\left(2 + \frac{1}{4}\right)} \cdot y^{\left(2 + \frac{3}{4}\right)}$[/tex]
5. [tex]$(81)^{\frac{1}{4}} \cdot(7)^{\frac{1}{4}} \cdot x^{\left(2 + \frac{1}{4}\right)} \cdot y^{\left(2 + \frac{\frac{3}{4}\right) + 3}{4}}$[/tex]
