Given that $\frac{\sqrt[3]{2^2} \times 3^{\frac{1}{5}}}{6^3} \times \left(\frac{2}{3}\right)^{-2} = 2^a \times 3^b$, find the fractions, $a$ and $b$. Given that $3^{x+2} \times 5^{x-1} = 9^x \times 5^{2x}$, evaluate $15^x$ without using a calculator. Given that $n$ is a positive integer, explain why the largest prime factor of $2(5^{n+1}) - 4(5^n) + 5^{n+2}$ is 31.

1. Express all terms with prime bases 2 and 3

$\frac{2^{\frac{2}{3}} \times 3^{\frac{1}{5}}}{(2 \times 3)^3} \times \frac{2^{-2}}{3^{-2}}$
$= \frac{2^{\frac{2}{3}} \times 3^{\frac{1}{5}}}{2^3 \times 3^3} \times 2^{-2} \times 3^2$

2. Apply laws of indices to combine the powers of 2 and 3

$= 2^{\frac{2}{3} - 3 - 2} \times 3^{\frac{1}{5} - 3 + 2}$
$= 2^{-\frac{13}{3}} \times 3^{-\frac{4}{5}}$

3. Compare exponents to find $a$ and $b$

Comparing with $2^a \times 3^b$:
$a = -\frac{13}{3}$
$b = -\frac{4}{5}$

1. Express all terms with prime bases and apply index laws to separate the terms with $x$.

$3^{x+2} \times 5^{x-1} = (3^2)^x \times 5^{2x}$
$3^x \cdot 3^2 \times 5^x \cdot 5^{-1} = 3^{2x} \times 5^{2x}$

2. Group terms with $x$ on one side and constants on the other.

$\frac{3^2}{5^1} = \frac{3^{2x} \times 5^{2x}}{3^x \times 5^x}$
$\frac{9}{5} = 3^{2x-x} \times 5^{2x-x}$
$\frac{9}{5} = 3^x \times 5^x$

3. Combine the terms with $x$ to form $15^x$.

$\frac{9}{5} = (3 \times 5)^x$
$15^x = \frac{9}{5}$

1. Factorise the common term $5^n$ from the expression.

$2(5^{n+1}) - 4(5^n) + 5^{n+2}$
$= 5^n(2 \times 5^1 - 4 + 5^2)$

2. Evaluate the numerical expression inside the bracket.

$= 5^n(10 - 4 + 25)$
$= 5^n(31)$

3. Explain why 31 is the largest prime factor.

Since $n$ is a positive integer, the prime factors of $31 \times 5^n$ are 5 and 31. Therefore, the largest prime factor is 31.