When written as the product of their prime factors, $p$ is $2^2\times3\times5$, $q$ is $2^6\times3^2$, $r$ is $2^2\times3^2\times11$. Find

1. Find the cube root from the prime factorisation of $q$.

Given $q=2^6\times3^2$,

$\sqrt[3]{q}=\sqrt[3]{2^6\times3^2}$

Only powers whose exponents are multiples of 3 can be taken out exactly:

$\sqrt[3]{2^6}=2^2$

However, $\sqrt[3]{3^2}$ is not an integer, so

$\sqrt[3]{q}=2^2\sqrt[3]{3^2}=4\sqrt[3]{9}$

This is not equal to $2^2\times3$.

1. Identify the highest power of each prime factor across $p$, $q$ and $r$.

$p=2^2\times3\times5$

$q=2^6\times3^2$

$r=2^2\times3^2\times11$

Highest powers: $2^6$, $3^2$, $5$, $11$.

2. Write the LCM as the product of its prime factors.

$\text{LCM}=2^6\times3^2\times5\times11$

1. Find the common prime factors with their lowest powers across $p$, $q$ and $r$.

$p=2^2\times3\times5$

$q=2^6\times3^2$

$r=2^2\times3^2\times11$

The common prime factors are $2$ and $3$.

For the HCF, take the lowest powers:

$2^2\times3^1$

2. Evaluate the HCF.

$2^2\times3=4\times3=12$