The expression shown in the image is a complex combination of nested radicals and exponents. Let me break it down and simplify step by step.

Given Expression:

[ 4 \cdot \sqrt[5]{\frac{486x^{17}y^{13}}{2050z^{18}}} \cdot 2 \cdot \left( \sqrt[4]{\frac{\sqrt{16x^8}}{y^{16}z^{20}}} \right) ]

Step-by-step Simplification:

1. Simplify the second term (inside the parenthesis): $\sqrt[4]{\frac{\sqrt{16x^8}}{y^{16}z^{20}}}$

   • First, simplify $\sqrt{16x^8}$.
     • $\sqrt{16} = 4$ and $\sqrt{x^8} = x^4$, so: $\sqrt{16x^8} = 4x^4$
   • Now, substitute this back into the expression: $\sqrt[4]{\frac{4x^4}{y^{16}z^{20}}}$
   • Take the fourth root of both numerator and denominator: [ \frac{\sqrt[4]{4x^4}}{\sqrt[4]{y^{16}z^{20}}} ]
     • $\sqrt[4]{4x^4} = 4^{1/4}x$ and $\sqrt[4]{y^{16}} = y^4$, $\sqrt[4]{z^{20}} = z^5$, so: $\frac{4^{1/4}x}{y^4z^5}$

2. Substitute back into the original expression: The expression now becomes: $4 \cdot \sqrt[5]{\frac{486x^{17}y^{13}}{2050z^{18}}} \cdot 2 \cdot \frac{4^{1/4}x}{y^4z^5}$

3. Combine constants and powers of $x, y, z$:

   • Combine the constant terms: $4 \cdot 2 \cdot 4^{1/4} = 8 \cdot 4^{1/4}$
   • Combine the powers of $x$, $y$, and $z$:
     • For $x$: $x^{17} \cdot x = x^{18}$
     • For $y$: $y^{13} / y^4 = y^9$
     • For $z$: $z^{18} \cdot z^5 = z^{23}$ So, the expression becomes: $8 \cdot 4^{1/4} \cdot \sqrt[5]{\frac{486x^{18}y^9}{2050z^{23}}}$

This is the simplified expression. Would you like further details on any step?

Follow-up questions:

1. What is the significance of fractional exponents and how can they be interpreted as roots?
2. How do you simplify expressions involving multiple radical layers?
3. How do you deal with radicals in the numerator and denominator when simplifying?
4. What are the rules for simplifying powers of variables in a radical expression?
5. How do constant terms behave when placed inside and outside radical expressions?

Tip:

When simplifying nested radical expressions, work step-by-step from the innermost radical outward to avoid confusion.