Let's solve the problem step-by-step:

(a) Simplify $y = x^{a-b} \times x^{b-c} \times x^{c-a} \times x^{-a-b}$

Using the laws of exponents, we combine the exponents since the base is the same ($x$):

$y = x^{(a-b) + (b-c) + (c-a) + (-a-b)}$

Now simplify the exponents:

1. Combine like terms for $a$, $b$, and $c$: $(a-b) + (b-c) + (c-a) + (-a-b) = a - b + b - c + c - a - a - b$

2. Cancel terms:

   • $a - a - a = -a$
   • $-b + b - b = -b$
   • $-c + c = 0$

This leaves: $y = x^{-a - b}$

Simplified result: $y = x^{-a-b}$

(b) Simplify $\left( \frac{16}{81} \right)^{\frac{3}{4}}$

Using the property of exponents for fractions: $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$

We first raise $16$ and $81$ to the power of $\frac{3}{4}$:

1. Rewrite $16$ and $81$ as powers of $2$ and $3$:
   • $16 = 2^4$
   • $81 = 3^4$

So: $\left( \frac{16}{81} \right)^{\frac{3}{4}} = \left( \frac{2^4}{3^4} \right)^{\frac{3}{4}}$

2. Apply the power rule $(a^m)^n = a^{m \cdot n}$:
   • $(2^4)^{\frac{3}{4}} = 2^{4 \cdot \frac{3}{4}} = 2^3 = 8$
   • $(3^4)^{\frac{3}{4}} = 3^{4 \cdot \frac{3}{4}} = 3^3 = 27$

Thus: $\left( \frac{16}{81} \right)^{\frac{3}{4}} = \frac{2^3}{3^3} = \frac{8}{27}$

Simplified result: $\frac{8}{27}$

Let me know if you'd like further clarification or additional steps! 😊

Related Questions:

1. Can you derive the laws of exponents for combining powers with the same base?
2. What happens if the exponents in part (a) were fractions instead of integers?
3. How would you handle a negative base in part (b)?
4. What is the general rule for simplifying fractional exponents like $\frac{3}{4}$?
5. How can the process in (b) be applied to roots and radicals?

Tip: Always simplify exponents step-by-step by grouping terms with the same base first!