To solve the given problem, we need to simplify the left-hand side and equate it to the given right-hand side $\frac{p^{a+b}}{q^{a-b}}$. We will then extract simultaneous equations to determine $a$ and $b$.

The Problem:

$\frac{\left( (2p^3q^4)^4 \div (-3q^5)^2 \right)}{9} \div \left( (4p^2q)^2 \right) = \frac{p^{a+b}}{q^{a-b}}.$

Step 1: Simplify the numerator

The numerator is: $\left( (2p^3q^4)^4 \div (-3q^5)^2 \right).$

• First, simplify $(2p^3q^4)^4$: $(2p^3q^4)^4 = 2^4 \cdot (p^3)^4 \cdot (q^4)^4 = 16p^{12}q^{16}.$
• Next, simplify $(-3q^5)^2$: $(-3q^5)^2 = (-3)^2 \cdot (q^5)^2 = 9q^{10}.$
• Now divide: $\frac{(2p^3q^4)^4}{(-3q^5)^2} = \frac{16p^{12}q^{16}}{9q^{10}} = \frac{16p^{12}}{9} \cdot q^{16-10} = \frac{16p^{12}q^6}{9}.$

Step 2: Divide by 9

Next, divide the result by $9$: $\frac{\frac{16p^{12}q^6}{9}}{9} = \frac{16p^{12}q^6}{81}.$

Step 3: Simplify the denominator

The denominator is: $(4p^2q)^2.$ Simplify: $(4p^2q)^2 = 4^2 \cdot (p^2)^2 \cdot q^2 = 16p^4q^2.$

Step 4: Combine numerator and denominator

Now divide the simplified numerator by the denominator: $\frac{\frac{16p^{12}q^6}{81}}{16p^4q^2}.$

• Simplify the constants: $\frac{16}{81} \div 16 = \frac{1}{81}.$
• Simplify the powers of $p$ and $q$: $p^{12} \div p^4 = p^{12-4} = p^8, \quad q^6 \div q^2 = q^{6-2} = q^4.$ Thus: $\frac{\frac{16p^{12}q^6}{81}}{16p^4q^2} = \frac{p^8q^4}{81}.$

Step 5: Equate to the given expression

We are given: $\frac{p^8q^4}{81} = \frac{p^{a+b}}{q^{a-b}}.$

• Comparing the powers of $p$: $a + b = 8.$
• Comparing the powers of $q$: $-(a - b) = 4 \quad \implies \quad a - b = -4.$

Step 6: Solve the simultaneous equations

We now solve:

1. $a + b = 8$,
2. $a - b = -4$.

Add the two equations: $(a + b) + (a - b) = 8 + (-4) \implies 2a = 4 \implies a = 2.$ Substitute $a = 2$ into $a + b = 8$: $2 + b = 8 \implies b = 6.$

Final Answer:

The values of $a$ and $b$ are: $\boxed{a = 2, \, b = 6.}$

Let me know if you need a detailed explanation of any step!

Here are 5 related questions for practice:

1. Simplify $(3p^2q)^3 \div (9pq^4)$.
2. Solve for $x$ and $y$ if $x + y = 10$ and $x - y = 2$.
3. Simplify $\frac{(2a^3b^2)^2}{4a^4b}$.
4. Evaluate $\frac{(p^2q^3)^4}{p^8q^{12}}$.
5. Solve $x^2 = 4^2 \div 2$.

Tip: When dividing powers of the same base, subtract the exponents!