Written as a product of its prime factors, $p = 3^3 \times 5 \times 7^2$.

1. Break 450 down into prime factors.

$450 = 45 \times 10$

$= (3 \times 3 \times 5) \times (2 \times 5)$

2. Write the prime factors in index form.

$450 = 2 \times 3 \times 3 \times 5 \times 5$

$= 2 \times 3^2 \times 5^2$

1. Compare the prime factorisations of $p$ and 450 and take the common factors with the lowest powers.

$p = 3^3 \times 5 \times 7^2$

From part (a),
$450 = 2 \times 3^2 \times 5^2$

Common prime factors: $3$ and $5$.

So the HCF is
$3^2 \times 5 = 45.$

1. Compare the prime powers in $450$ with those needed for $p$.

Given
$p = 3^3 \times 5 \times 7^2$
and from part (a),
$450 = 2 \times 3^2 \times 5^2.$
For $450m$ to be a multiple of $p$, it must contain at least $3^3 \times 5 \times 7^2$.

$450$ already has $3^2$ and $5^2$, so it still needs one more factor of $3$ and two factors of $7$.

2. Find the smallest possible value of $m$.

So the smallest integer $m$ is
$m = 3 \times 7^2 = 147.$

1. Identify how the prime powers in $p$ must change to become a perfect cube.

Given
$p=3^3\times 5\times 7^2$.

For a perfect cube, every prime factor must have an exponent that is a multiple of $3$.

$3^3$ is already suitable.

$5^1$ should be removed by dividing by $5$.

$7^2$ should be multiplied by $7$ to become $7^3$.

2. Choose the prime numbers $a$ and $b$ and verify the product is a perfect cube.

Hence
$\frac{a}{b}=\frac{7}{5}.$

Then
$p\times \frac{7}{5}=3^3\times 5\times 7^2\times \frac{7}{5}=3^3\times 7^3,$
which is a perfect cube.

So $a=7$ and $b=5$.